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stringlengths 2
88
| description
stringlengths 31
8.62k
| public_tests
dict | private_tests
dict | solution_type
stringclasses 2
values | programming_language
stringclasses 5
values | solution
stringlengths 1
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1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.io.*;
public class SummoningMinions {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
int t = Integer.parseInt(br.readLine());
Comparator<int[]> comp = (a, b) -> a[2] - b[2];
while (--t >= 0) {
String[] line = br.readLine().split(" ");
int n = Integer.parseInt(line[0]), k = Integer.parseInt(line[1]);
int[][] ab = new int[n][3];
for (int i = 0; i < n; ++i) {
line = br.readLine().split(" ");
ab[i][0] = i;
ab[i][1] = Integer.parseInt(line[0]);
ab[i][2] = Integer.parseInt(line[1]);
}
Arrays.sort(ab, comp);
int[][] dp = new int[n+1][n+1];
for (int i = 1; i <= n; ++i) {
dp[i][0] = dp[i-1][0] + ab[i-1][2] * (k - 1);
for (int j = 1; j < i; ++j) {
dp[i][j] = Math.max(dp[i-1][j-1] + ab[i-1][2] * (j - 1) + ab[i-1][1], dp[i-1][j] + ab[i-1][2] * (k - 1));
}
dp[i][i] = dp[i-1][i-1] + ab[i-1][2] * (i - 1) + ab[i-1][1];
}
int[] res = new int[n*2-k];
int j1 = k, j2 = k - 1;
for (int i = n; i > 0; --i) {
if (i > j1 && dp[i][j1] == dp[i-1][j1] + ab[i-1][2] * (k - 1)) {
res[++j2] = ab[i-1][0];
} else {
res[--j1] = ab[i-1][0];
}
}
bw.write(res.length + '0');
bw.newLine();
for (int i = 0; i < k - 1; ++i) {
bw.write(res[i] + '1');
bw.write(' ');
}
for (int i = k; i < n; ++i) {
bw.write(res[i] + '1');
bw.write(' ');
bw.write('-');
bw.write(res[i] + '1');
bw.write(' ');
}
bw.write(res[k-1] + '1');
bw.newLine();
}
br.close();
bw.close();
}
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double eps = 1e-12;
const int inf = 2000000000;
const long long int infLL = (long long int)1e18;
long long int MOD = 1000000007;
int MOD1 = 1000000007;
int MOD2 = 1000000009;
inline bool checkBit(long long int n, long long int i) {
return n & (1LL << i);
}
inline long long int setBit(long long int n, long long int i) {
return n | (1LL << i);
;
}
inline long long int resetBit(long long int n, long long int i) {
return n & (~(1LL << i));
}
int dx[] = {0, 0, +1, -1};
int dy[] = {+1, -1, 0, 0};
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long int year) {
return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline void normal(long long int &a) {
a %= MOD;
(a < 0) && (a += MOD);
}
inline long long int modMul(long long int a, long long int b) {
a %= MOD, b %= MOD;
normal(a), normal(b);
return (a * b) % MOD;
}
inline long long int modAdd(long long int a, long long int b) {
a %= MOD, b %= MOD;
normal(a), normal(b);
return (a + b) % MOD;
}
inline long long int modSub(long long int a, long long int b) {
a %= MOD, b %= MOD;
normal(a), normal(b);
a -= b;
normal(a);
return a;
}
inline long long int modPow(long long int b, long long int p) {
long long int r = 1LL;
while (p) {
if (p & 1) r = modMul(r, b);
b = modMul(b, b);
p >>= 1LL;
}
return r;
}
inline long long int modDiv(long long int a, long long int b) {
return modMul(a, modPow(b, MOD - 2));
}
bool comp(const pair<long long int, pair<long long int, long long int> > &p1,
const pair<long long int, pair<long long int, long long int> > &p2) {
return p1.first > p2.first;
}
bool comp1(const pair<long long int, long long int> &p1,
const pair<long long int, long long int> &p2) {
if (p1.first == p2.first) {
return p1.second > p2.second;
}
return p1.first < p2.first;
}
long long int converter(string a) {
long long int i, mul = 1LL, r, t, ans = 0LL;
if (a.length() == 0) return 0;
for (i = a.length() - 1; i >= 0; i--) {
t = a[i] - '0';
r = t % 10;
ans += (mul * r);
mul = mul * 10;
}
return ans;
}
int msb(unsigned x) {
union {
double a;
int b[2];
};
a = x;
return (b[1] >> 20) - 1023;
}
const int MAX = 78;
int t, n, k;
pair<pair<long long int, long long int>, long long int> p[MAX];
long long int dp[MAX][MAX];
vector<long long int> good, bad;
long long int solve(int idx, int cnt) {
if (idx == n + 1) {
if (cnt != k) return -infLL;
return 0;
}
if (dp[idx][cnt] != -1) return dp[idx][cnt];
long long int ret = 0;
if (cnt + 1 <= k) {
ret = max(ret, cnt * p[idx].first.first + p[idx].first.second +
solve(idx + 1, cnt + 1));
}
ret = max(ret, ((k - 1) * p[idx].first.first) + solve(idx + 1, cnt));
return dp[idx][cnt] = ret;
}
void trace(int idx, int cnt) {
if (idx == n + 1) return;
long long int ret1 = 0LL, ret2 = 0LL;
if (cnt + 1 <= k)
ret1 = cnt * p[idx].first.first + p[idx].first.second +
solve(idx + 1, cnt + 1);
ret2 = ((k - 1) * p[idx].first.first) + solve(idx + 1, cnt);
if (ret1 >= ret2) {
trace(idx + 1, cnt + 1);
good.push_back(p[idx].second);
} else {
trace(idx + 1, cnt);
bad.push_back(p[idx].second);
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
cout.unsetf(ios::floatfield);
cout.precision(20);
cout.setf(ios::fixed, ios::floatfield);
;
cin >> t;
while (t--) {
good.clear();
bad.clear();
long long int maxi = 0, idx;
cin >> n >> k;
for (int i = 1; i <= n; ++i) {
cin >> p[i].first.second >> p[i].first.first;
p[i].second = i;
if (p[i].first.second >= maxi) maxi = p[i].first.second, idx = i;
}
if (k == 1) {
cout << 1 << '\n';
cout << idx << '\n';
continue;
}
sort(p + 1, p + n + 1);
memset(dp, -1, sizeof(dp));
long long int now = solve(1, 0);
trace(1, 0);
reverse((good).begin(), (good).end());
reverse((bad).begin(), (bad).end());
cout << good.size() + (2 * bad.size()) << '\n';
for (int i = 0; i < (int)good.size() - 1; ++i) {
cout << good[i] << " ";
}
for (int i = 0; i < bad.size(); ++i) {
cout << bad[i] << " " << -1 * bad[i] << " ";
}
if (good.size() > 0) cout << good[good.size() - 1] << '\n';
}
return 0;
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
python3
|
def m(e):
return e['m']
def h(e):
return e['h']
a = int(input())
for i in range(a):
num = [int(x) for x in input().split()]
mi = []
for j in range(num[0]):
a = [int(x) for x in input().split()]
k = {}
k['m'] = a[0]
k['h'] = a[1]
k['i'] = j+1
mi.append(k)
mi.sort(key=m,reverse = True)
cho = mi
nch = mi
cho = mi[:num[1]]
nch = mi[num[1]:]
klen = 0
k = []
for element in cho[:num[1]-1]:
#print(element['i'],end = " ")
k.append(element['i'])
klen +=1
#print()
#print(nch)
if(num[1]!=1):
for element in nch:
if(element['h']!=0):
#print(element['i'],end = " ")
#print(-1*element['i'],end = " ")
k.append(element['i'])
klen +=1
k.append(-1*element['i'])
klen +=1
k.append(cho[num[1]-1]['i'])
klen+=1
#print(cho[num[1]-1]['i'],end = " ")
print(klen)
for j in k:
print(j,end = " ")
print()
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
void *wmem;
char memarr[96000000];
template <class T>
inline void walloc1d(T **arr, int x, void **mem = &wmem) {
static int skip[16] = {0, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1};
(*mem) = (void *)(((char *)(*mem)) + skip[((unsigned long long)(*mem)) & 15]);
(*arr) = (T *)(*mem);
(*mem) = ((*arr) + x);
}
template <class T1, class T2>
void sortA_L(int N, T1 a[], T2 b[], void *mem = wmem) {
int i;
pair<T1, T2> *arr;
walloc1d(&arr, N, &mem);
for (i = (0); i < (N); i++) {
arr[i].first = a[i];
arr[i].second = b[i];
}
sort(arr, arr + N);
for (i = (0); i < (N); i++) {
a[i] = arr[i].first;
b[i] = arr[i].second;
}
}
template <class T1, class T2, class T3>
void sortA_L(int N, T1 a[], T2 b[], T3 c[], void *mem = wmem) {
int i;
pair<T1, pair<T2, T3> > *arr;
walloc1d(&arr, N, &mem);
for (i = (0); i < (N); i++) {
arr[i].first = a[i];
arr[i].second.first = b[i];
arr[i].second.second = c[i];
}
sort(arr, arr + N);
for (i = (0); i < (N); i++) {
a[i] = arr[i].first;
b[i] = arr[i].second.first;
c[i] = arr[i].second.second;
}
}
inline int my_getchar() {
static char buf[1048576];
static int s = 1048576;
static int e = 1048576;
if (s == e && e == 1048576) {
e = fread(buf, 1, 1048576, stdin);
s = 0;
}
if (s == e) {
return EOF;
}
return buf[s++];
}
inline void rd(int &x) {
int k;
int m = 0;
x = 0;
for (;;) {
k = my_getchar();
if (k == '-') {
m = 1;
break;
}
if ('0' <= k && k <= '9') {
x = k - '0';
break;
}
}
for (;;) {
k = my_getchar();
if (k < '0' || k > '9') {
break;
}
x = x * 10 + k - '0';
}
if (m) {
x = -x;
}
}
inline void rd(long long &x) {
int k;
int m = 0;
x = 0;
for (;;) {
k = my_getchar();
if (k == '-') {
m = 1;
break;
}
if ('0' <= k && k <= '9') {
x = k - '0';
break;
}
}
for (;;) {
k = my_getchar();
if (k < '0' || k > '9') {
break;
}
x = x * 10 + k - '0';
}
if (m) {
x = -x;
}
}
inline int rd_int(void) {
int x;
rd(x);
return x;
}
struct MY_WRITER {
char buf[1048576];
int s;
int e;
MY_WRITER() {
s = 0;
e = 1048576;
}
~MY_WRITER() {
if (s) {
fwrite(buf, 1, s, stdout);
}
}
};
MY_WRITER MY_WRITER_VAR;
void my_putchar(int a) {
if (MY_WRITER_VAR.s == MY_WRITER_VAR.e) {
fwrite(MY_WRITER_VAR.buf, 1, MY_WRITER_VAR.s, stdout);
MY_WRITER_VAR.s = 0;
}
MY_WRITER_VAR.buf[MY_WRITER_VAR.s++] = a;
}
inline void wt_L(char a) { my_putchar(a); }
inline void wt_L(int x) {
int s = 0;
int m = 0;
char f[10];
if (x < 0) {
m = 1;
x = -x;
}
while (x) {
f[s++] = x % 10;
x /= 10;
}
if (!s) {
f[s++] = 0;
}
if (m) {
my_putchar('-');
}
while (s--) {
my_putchar(f[s] + '0');
}
}
int N;
int K;
int ind[75];
long long A[75];
long long B[75];
long long dp[76][76];
int bk[76][76];
int use[75];
int fg[75];
int ress;
int res[150];
int main() {
int Lj4PdHRW;
wmem = memarr;
int KL2GvlyY = rd_int();
for (Lj4PdHRW = (0); Lj4PdHRW < (KL2GvlyY); Lj4PdHRW++) {
int i;
int j;
rd(N);
rd(K);
{
int cTE1_r3A;
for (cTE1_r3A = (0); cTE1_r3A < (N); cTE1_r3A++) {
rd(A[cTE1_r3A]);
rd(B[cTE1_r3A]);
}
}
for (i = (0); i < (N); i++) {
ind[i] = i + 1;
}
sortA_L(N, B, A, ind);
for (i = (0); i < (N + 1); i++) {
for (j = (0); j < (N + 1); j++) {
dp[i][j] = -4611686016279904256LL;
}
}
dp[0][0] = 0;
for (i = (0); i < (N); i++) {
for (j = (0); j < (i + 1); j++) {
if (dp[i][j] > -4611686016279904256LL) {
if (dp[i + 1][j] < dp[i][j] + B[i] * (K - 1)) {
dp[i + 1][j] = dp[i][j] + B[i] * (K - 1);
bk[i + 1][j] = j;
}
if (dp[i + 1][j + 1] < dp[i][j] + A[i] + B[i] * j) {
dp[i + 1][j + 1] = dp[i][j] + A[i] + B[i] * j;
bk[i + 1][j + 1] = j;
}
}
}
}
j = K;
for (i = (N)-1; i >= (0); i--) {
if (bk[i + 1][j] == j) {
continue;
}
j--;
use[j] = i;
fg[i] = 1;
}
ress = 0;
for (i = (0); i < (K - 1); i++) {
res[ress++] = ind[use[i]];
}
for (i = (0); i < (N); i++) {
if (!fg[i]) {
res[ress++] = ind[i];
res[ress++] = -ind[i];
}
}
res[ress++] = ind[use[K - 1]];
wt_L(ress);
wt_L('\n');
{
int XJIcIBrW;
if (ress == 0) {
wt_L('\n');
} else {
for (XJIcIBrW = (0); XJIcIBrW < (ress - 1); XJIcIBrW++) {
wt_L(res[XJIcIBrW]);
wt_L(' ');
}
wt_L(res[XJIcIBrW]);
wt_L('\n');
}
}
}
return 0;
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
java
|
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;
/**
*
* @author MasterEx
*/
public class ECR87F {
static class Minion {
int pos;
long power;
long benefit;
public Minion(int p, long po, long b) {
pos = p + 1;
power = po;
benefit = b;
}
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
for (int i = 0; i < t; i++) {
int n = in.nextInt();
int k = in.nextInt();
LinkedList<Minion> minions = new LinkedList<>();
for (int j = 0; j < n; j++) {
long a = in.nextInt();
long b = in.nextInt();
minions.add(new Minion(j, a, b));
}
minions.sort((Minion m1, Minion m2) -> {
if (m1.power == m2.power) {
return Long.compare(m2.benefit, m1.benefit);
}
return Long.compare(m2.power, m1.power);
});
String s = "";
int m = 0;
for (int j = 0; j < k - 1 && !minions.isEmpty(); j++) {
m++;
s += minions.removeFirst().pos + " ";
}
if (!minions.isEmpty()) {
minions.sort((m1, m2) -> {
if (m1.benefit == m2.benefit) {
return Long.compare(m2.power, m1.power);
}
return Long.compare(m2.benefit, m1.benefit);
});
}
while (!minions.isEmpty() && m <= 2 * n) {
Minion minion = minions.removeFirst();
m++;
s += minion.pos;
if (minion.benefit == 0
|| !minions.isEmpty() && minion.benefit == minions.peek().benefit) {
break;
}
if (!minions.isEmpty()) {
m++;
s += " -" + minion.pos + " ";
}
}
System.out.println(m);
System.out.println(s);
}
}
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
python3
|
for _ in range(int(input())):
n, k = map(int, input().split())
values = []
for i in range(n):
a, b = map(int, input().split())
values.append((i + 1, a, b))
if k == 1:
best = -1
score = 0
for (i, a, b) in values:
if a > score:
best = i
score = a
print(best)
else:
values = sorted(values, key=lambda x: x[1], reverse=True)
keepers = sorted(values[:k], key=lambda x: x[2])
destroyers = sorted([x for x in values[k:] if x[2] > 0], key=lambda x: x[2])
if len(destroyers):
turns = []
for i, a, b in keepers[:-1]:
turns.append(i)
for i, a, b in destroyers:
turns.append(i)
turns.append(-i)
turns.append(keepers[-1][0])
else:
turns = [i for i,a, b in keepers]
print(len(turns))
print(*turns)
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
vector<pair<pair<int, int>, int>> v;
long long dp[80][80];
pair<int, int> from[80][80];
set<int> s;
int main() {
int t;
cin >> t;
while (t) {
t--;
int i, j, n, k;
cin >> n >> k;
v.clear();
s.clear();
for (i = 1; i <= n; i++) {
int a, b;
cin >> a >> b;
a -= b;
v.push_back({{b, a}, i});
s.insert(i);
}
sort(v.rbegin(), v.rend());
for (i = 0; i <= n; i++) {
for (j = 0; j <= k; j++) {
dp[i][j] = -100000000000;
from[i][j] = {-1, -1};
}
}
dp[0][0] = 0;
for (i = 0; i < n; i++) {
int ind = v[i].second;
long long a = v[i].first.second, b = v[i].first.first;
if (dp[i + 1][0] < dp[i][0]) {
from[i + 1][0] = {ind, 0};
dp[i + 1][0] = dp[i][0];
}
for (j = 1; j <= k; j++) {
if (dp[i + 1][j] < dp[i][j]) {
from[i + 1][j] = {ind, j};
dp[i + 1][j] = dp[i][j];
}
if (dp[i + 1][j] < dp[i][j - 1] + (k - j) * b + a) {
from[i + 1][j] = {ind, j - 1};
dp[i + 1][j] = dp[i][j - 1] + (k - j) * b + a;
}
}
}
vector<int> ans;
int nx = n, ny = k;
while (ny != 0) {
pair<int, int> p = from[nx][ny];
if (p.second != ny) {
ans.push_back(p.first);
s.erase(p.first);
}
nx--;
ny = p.second;
}
cout << 2 * n - k << endl;
for (i = 0; i < k - 1; i++) {
cout << ans[i] << " ";
}
for (int x : s) {
cout << x << " " << -x << " ";
}
cout << ans[k - 1] << endl;
}
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
cpp
|
///#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
///#pragma GCC target("sse,sse2,sse3,ssse3,sse4,abm,mmx,popcnt,avx,avx2,tune=native")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
/*
zet:
find_by_order(k): k-stat [k > 0]
order_of_key(k) : {el < k}
gp_hash_table<key, val, custom_hash>
rope<T>:
iterators: mutable_begin, ..
methods : erase, insert, substr, ..
*/
#define all(x) (x).begin(), (x).end()
#define size(x) (int)((x).size())
#define em_back emplace_back
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template<typename T>
using zet = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
struct custom_hash {
static uint64_t splitmix64(uint64_t x) {
// http://xorshift.di.unimi.it/splitmix64.c
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
return splitmix64(x + FIXED_RANDOM);
}
};
mt19937 eng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 2 * 75 + 4;
struct edge {
int u, v, w, c, f;
edge(int u_, int v_, int w_, int c_, int f_) :
u(u_), v(v_), w(w_), c(c_), f(f_) {}
};
vector<edge> edges;
vector<int> g[N];
int n, k, s, t, a[N], b[N];
void read() {
cin >> n >> k;
for (int i = 1; i <= n; ++i) {
cin >> a[i] >> b[i];
}
}
void add(int u, int v, int w) {
int sz = size(edges);
edges.em_back(u, v, w, 1, 0);
edges.em_back(v, u, -w, 0, 0);
g[u].em_back(sz);
g[v].em_back(sz ^ 1);
}
void prepare() {
edges.clear();
s = 0, t = 2 * n + 1;
for (int i = 1; i <= n; ++i) {
add(s, i, 0);
add(n + i, t, 0);
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
int l = i, r = j + n;
int cost = 0;
if (j == n) {
cost = a[i] + b[i] * (k - 1);
} else if (j < k) {
cost = a[i] + b[i] * j;
} else {
cost = b[i] * (k - 1);
}
add(l, r, -cost);
}
}
}
const int INF = 1e9 + 4;
ll d[N], phi[N];
int p[N];
void bellmanFord() {
fill_n(phi, N, INF);
phi[s] = 0;
for (int i = 0; i < t - 1; ++i) {
for (edge& e : edges) {
if (e.c > 0 && phi[e.u] < INF) {
phi[e.v] = min(phi[e.v], phi[e.u] + e.w);
}
}
}
}
bool dijk() {
memset(p, -1, sizeof p);
vector<bool> used(t + 4);
fill(d, d + N, INF);
d[s] = 0;
while (true) {
int u = -1;
for (int v = s; v <= t; ++v) {
if (!used[v] && (u == -1 || d[v] < d[u])) {
u = v;
}
}
if (u == -1 || d[u] == INF) {
break;
}
used[u] = true;
for (int num : g[u]) {
edge& e = edges[num];
ll len = e.w + phi[u] - phi[e.v];
if (e.f < e.c && d[u] + len < d[e.v]) {
d[e.v] = d[u] + len;
p[e.v] = num;
}
}
}
return p[t] != -1;
}
void mincost() {
bellmanFord();
while (dijk()) {
for (int i = s; i <= t; ++i) {
phi[i] += d[i];
}
int cmin = INF;
for (int v = t; v != s;) {
edge& e = edges[p[v]];
cmin = min(cmin, e.c - e.f);
v = e.u;
}
for (int v = t; v != s;) {
edge& e = edges[p[v]];
edge& re = edges[p[v] ^ 1];
e.f += cmin;
re.f -= cmin;
v = e.u;
}
}
}
vector<int> op;
vector<int> stonks;
void getStonks(int num) {
if (size(stonks) < k) {
op.em_back(num);
stonks.em_back(num);
} else {
if (size(stonks)) {
op.em_back(-stonks.back());
}
op.em_back(num);
if (size(stonks)) {
stonks.pop_back();
}
stonks.em_back(num);
}
}
bool check(int u, int v) {
for (int num : g[u]) {
edge& e = edges[num];
if (e.v == v && e.f > 0 && e.c > 0) {
return true;
}
}
return false;
}
void calc() {
op.clear();
stonks.clear();
for (int j = 1; j <= n; ++j) {
for (int i = 1; i <= n; ++i) {
int l = i, r = j + n;
if (check(l, r)) {
getStonks(i);
break;
}
}
}
}
void write() {
cout << size(op) << "\n";
for (int i : op) {
cout << i << " ";
}
cout << "\n";
}
void solve() {
read();
prepare();
mincost();
calc();
write();
}
int main() {
ios::sync_with_stdio(0);
cout.tie(0), cin.tie(0);
int z = 1;
cin >> z;
while (z--) {
solve();
}
return 0;
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double eps = 1e-12;
const int inf = 2000000000;
const long long int infLL = (long long int)1e18;
long long int MOD = 1000000007;
int MOD1 = 1000000007;
int MOD2 = 1000000009;
inline bool checkBit(long long int n, long long int i) {
return n & (1LL << i);
}
inline long long int setBit(long long int n, long long int i) {
return n | (1LL << i);
;
}
inline long long int resetBit(long long int n, long long int i) {
return n & (~(1LL << i));
}
int dx[] = {0, 0, +1, -1};
int dy[] = {+1, -1, 0, 0};
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long int year) {
return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline void normal(long long int &a) {
a %= MOD;
(a < 0) && (a += MOD);
}
inline long long int modMul(long long int a, long long int b) {
a %= MOD, b %= MOD;
normal(a), normal(b);
return (a * b) % MOD;
}
inline long long int modAdd(long long int a, long long int b) {
a %= MOD, b %= MOD;
normal(a), normal(b);
return (a + b) % MOD;
}
inline long long int modSub(long long int a, long long int b) {
a %= MOD, b %= MOD;
normal(a), normal(b);
a -= b;
normal(a);
return a;
}
inline long long int modPow(long long int b, long long int p) {
long long int r = 1LL;
while (p) {
if (p & 1) r = modMul(r, b);
b = modMul(b, b);
p >>= 1LL;
}
return r;
}
inline long long int modDiv(long long int a, long long int b) {
return modMul(a, modPow(b, MOD - 2));
}
bool comp(const pair<long long int, pair<long long int, long long int> > &p1,
const pair<long long int, pair<long long int, long long int> > &p2) {
return p1.first > p2.first;
}
bool comp1(const pair<long long int, long long int> &p1,
const pair<long long int, long long int> &p2) {
if (p1.first == p2.first) {
return p1.second > p2.second;
}
return p1.first < p2.first;
}
long long int converter(string a) {
long long int i, mul = 1LL, r, t, ans = 0LL;
if (a.length() == 0) return 0;
for (i = a.length() - 1; i >= 0; i--) {
t = a[i] - '0';
r = t % 10;
ans += (mul * r);
mul = mul * 10;
}
return ans;
}
int msb(unsigned x) {
union {
double a;
int b[2];
};
a = x;
return (b[1] >> 20) - 1023;
}
const int MAX = 78;
int t, n, k;
pair<pair<long long int, long long int>, long long int> p[MAX];
long long int dp[MAX][MAX];
vector<long long int> good, bad;
long long int solve(int idx, int cnt) {
if (idx == n + 1) {
if (cnt != k) return -infLL;
return 0;
}
if (dp[idx][cnt] != -1) return dp[idx][cnt];
long long int ret = 0;
if (cnt + 1 <= k) {
ret = max(ret, cnt * p[idx].first.first + p[idx].first.second +
solve(idx + 1, cnt + 1));
}
ret = max(ret, ((k - 1) * p[idx].first.first) + solve(idx + 1, cnt));
return dp[idx][cnt] = ret;
}
void trace(int idx, int cnt) {
if (idx == n + 1) return;
long long int ret1 = 0LL, ret2 = 0LL;
if (cnt + 1 <= k)
ret1 = cnt * p[idx].first.first + p[idx].first.second +
solve(idx + 1, cnt + 1);
ret2 = ((k - 1) * p[idx].first.first) + solve(idx + 1, cnt);
if (ret1 > ret2) {
trace(idx + 1, cnt + 1);
good.push_back(p[idx].second);
} else {
trace(idx + 1, cnt);
bad.push_back(p[idx].second);
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
cout.unsetf(ios::floatfield);
cout.precision(20);
cout.setf(ios::fixed, ios::floatfield);
;
cin >> t;
while (t--) {
good.clear();
bad.clear();
long long int maxi = 0, idx;
cin >> n >> k;
for (int i = 1; i <= n; ++i) {
cin >> p[i].first.second >> p[i].first.first;
p[i].second = i;
if (p[i].first.second >= maxi) maxi = p[i].first.second, idx = i;
}
if (k == 1) {
cout << 1 << '\n';
cout << idx << '\n';
continue;
}
sort(p + 1, p + n + 1);
memset(dp, -1, sizeof(dp));
long long int now = solve(1, 0);
;
trace(1, 0);
reverse((good).begin(), (good).end());
reverse((bad).begin(), (bad).end());
if (good.size() == 0) {
cout << 1000000000000 + k << '\n';
continue;
}
cout << good.size() + (2 * bad.size()) << '\n';
for (int i = 0; i < (int)good.size() - 1; ++i) {
cout << good[i] << " ";
}
for (int i = 0; i < bad.size(); ++i) {
cout << bad[i] << " " << -1 * bad[i] << " ";
}
cout << good[good.size() - 1] << '\n';
}
return 0;
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
cpp
|
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
typedef tree<long long,null_type,less<long long>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
typedef tree<long long,null_type,greater<long long>,rb_tree_tag,tree_order_statistics_node_update> ordered_set1;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
//Required Functions
ll power(ll x,ll y)
{
if (y == 0)
return 1;
ll p = power(x, y/2) % MOD;
p = (p * p) % MOD;
return (y%2 == 0)? p : (x * p) % MOD;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen ("INPUT.txt" , "r" , stdin);
//freopen ("OUTPUT.txt" , "w" , stdout);
}
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t clk;
clk = clock();
//-----------------------------------------------------------------------------------------------------------//
READV(T);
while(T--)
{
READV(N);
READV(K);
pair<pll,ll> Z[N];
FORI(i,0,N)
{
READ(Z[i].fi.se);
READ(Z[i].fi.fi);
Z[i].se=i+1;
}
sort(Z,Z+N);
ll dp[N+1][N+1];
FORI(i,0,N+1)
{
FORI(j,0,N+1)
{
dp[i][j]=-inf;
}
}
dp[0][0]=0;
ll back_tracker[N+1][N+1];
memset(back_tracker,0,sizeof(back_tracker));
FORI(i,0,N)
{
FORI(j,0,N)
{
if(dp[i][j]>=0)
{
if(dp[i+1][j]<dp[i][j]+j*Z[i].fi.fi)
{
dp[i+1][j]=dp[i][j]+j*Z[i].fi.fi;
back_tracker[i+1][j]=j;
}
if(dp[i+1][j+1]<dp[i][j]+Z[i].fi.se+j*Z[i].fi.fi)
{
dp[i+1][j+1]=dp[i][j]+Z[i].fi.se+j*Z[i].fi.fi;
back_tracker[i+1][j+1]=j;
}
}
}
}
vector<ll> ans1,ans2;
ll current=K;
FORD(i,N,0)
{
if(back_tracker[i][current]==current)
{
ans2.pb(Z[i-1].se);
}
else
{
ans1.pb(Z[i-1].se);
}
current=back_tracker[i][current];
}
reverse(ans1.begin(),ans1.end());
reverse(ans2.begin(),ans2.end());
cout<<sz(ans2)*2+sz(ans1)<<endl;
FORI(i,0,sz(ans1)-1)
{
cout<<ans1[i]<<" ";
}
FORI(i,0,sz(ans2))
{
cout<<ans2[i]<<" "<<-ans2[i]<<" ";
}
cout<<ans1.back()<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
clk = clock() - clk;
cerr << fixed << setprecision(6) << "Time: " << ((double)clk)/CLOCKS_PER_SEC << endl;
return 0;
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
python3
|
#qn given an arr of size n for every i position find a min j such that j>i and arr[j]>arr[i]
#it prints the position of the nearest .
import sys
input = sys.stdin.readline
#qn given an arr of size n for every i position find a min j such that j>i and arr[j]>arr[i]
#it prints the position of the nearest .
# import sys
import heapq
import copy
import math
#heapq.heapify(li)
#
#heapq.heappush(li,4)
#
#heapq.heappop(li)
#
# & Bitwise AND Operator 10 & 7 = 2
# | Bitwise OR Operator 10 | 7 = 15
# ^ Bitwise XOR Operator 10 ^ 7 = 13
# << Bitwise Left Shift operator 10<<2 = 40
# >> Bitwise Right Shift Operator
'''############ ---- Input Functions ---- #######Start#####'''
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
############ ---- Input Functions ---- #######End
# #####
class Node:
def _init_(self,val):
self.data = val
self.left = None
self.right = None
##to initialize wire : object_name= node(val)##to create a new node
## can also be used to create linked list
class fen_tree:
"""Implementation of a Binary Indexed Tree (Fennwick Tree)"""
#def __init__(self, list):
# """Initialize BIT with list in O(n*log(n))"""
# self.array = [0] * (len(list) + 1)
# for idx, val in enumerate(list):
# self.update(idx, val)
def __init__(self, list):
""""Initialize BIT with list in O(n)"""
self.array = [0] + list
for idx in range(1, len(self.array)):
idx2 = idx + (idx & -idx)
if idx2 < len(self.array):
self.array[idx2] += self.array[idx]
def prefix_query(self, idx):
"""Computes prefix sum of up to including the idx-th element"""
# idx += 1
result = 0
while idx:
result += self.array[idx]
idx -= idx & -idx
return result
def prints(self):
print(self.array)
return
# for i in self.array:
# print(i,end = " ")
# return
def range_query(self, from_idx, to_idx):
"""Computes the range sum between two indices (both inclusive)"""
return self.prefix_query(to_idx) - self.prefix_query(from_idx - 1)
def update(self, idx, add):
"""Add a value to the idx-th element"""
# idx += 1
while idx < len(self.array):
self.array[idx] += add
idx += idx & -idx
def pre_sum(arr):
#"""returns the prefix sum inclusive ie ith position in ans represent sum from 0 to ith position"""
p = [0]
for i in arr:
p.append(p[-1] + i)
p.pop(0)
return p
def pre_back(arr):
#"""returns the prefix sum inclusive ie ith position in ans represent sum from 0 to ith position"""
p = [0]
for i in arr:
p.append(p[-1] + i)
p.pop(0)
return p
def bin_search(arr,l,r,val):#strickly greater
if arr[r] <= val:
return r+1
if r-l < 2:
if arr[l]>val:
return l
else:
return r
mid = int((l+r)/2)
if arr[mid] <= val:
return bin_search(arr,mid,r,val)
else:
return bin_search(arr,l,mid,val)
def search_leftmost(arr,val):
def helper(arr,l,r,val):
# print(arr)
print(l,r)
if arr[l] == val:
return l
if r -l <=1:
if arr[r] == val:
return r
else:
print("not found")
return
mid = int((r+l)/2)
if arr[mid] >= val:
return helper(arr,l,mid,val)
else:
return helper(arr,mid,r,val)
return helper(arr,0,len(arr)-1,val)
def search_rightmost(arr,val):
def helper(arr,l,r,val):
# print(arr)
print(l,r)
if arr[r] == val:
return r
if r -l <=1:
if arr[l] == val:
return r
else:
print("not found")
return
mid = int((r+l)/2)
if arr[mid] > val:
return helper(arr,l,mid,val)
else:
return helper(arr,mid,r,val)
return helper(arr,0,len(arr)-1,val)
def pr_list(a):
print(*a, sep=" ")
def main():
tests = inp()
# tests = 1
mod = 1000000007
limit = 10**18
# print(limit)
for test in range(tests):
[n,k] = inlt()
a = [ 0 for i in range(n)]
b = [0 for i in range(n)]
for i in range(n):
[a[i],b[i]] = inlt()
if k == 1:
print(1)
maxi = max(a)
print(a.index(maxi) +1)
continue
out = []
power = [[b[i],a[i],i+1] for i in range(n)]
power.sort()
grp = []
k = min(n,k)
for i in range(k-1):
for j in range(len(grp)):
grp[j][0] = grp[j][0] + power[i][0]
grp.append([power[i][1],power[i][2]])
out.append(power[i][2])
grp.sort()
for i in range(k-1,n,1):
for j in range(len(grp)):
grp[j][0] = grp[j][0] + power[i][0]
grp.append([power[i][1],power[i][2]])
out.append(power[i][2])
grp.sort()
if i != n-1:
dum = grp.pop(0)
out.append(-1*dum[1])
print(len(out))
pr_list(out)
if __name__== "__main__":
main()
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
java
|
/*
ID: tommatt1
LANG: JAVA
TASK:
*/
import java.util.*;
import java.io.*;
public class cf1354f{
public static void main(String[] args)throws IOException {
PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer(bf.readLine());
int tt=Integer.parseInt(st.nextToken());
while(tt-->0) {
st=new StringTokenizer(bf.readLine());
int n=Integer.parseInt(st.nextToken());
int k=Integer.parseInt(st.nextToken());
pair[] mnn=new pair[n+1];
boolean[] inc=new boolean[n+1];
boolean[][] poss=new boolean[n+1][k+1];
int[][] dp=new int[n+1][k+1];
for(int i=1;i<=n;i++) {
st=new StringTokenizer(bf.readLine());
int a=Integer.parseInt(st.nextToken());
int b=Integer.parseInt(st.nextToken());
mnn[i]=new pair(a,b,i);
}
for(int i=0;i<=n;i++) {
Arrays.fill(dp[i], 1, k+1, Integer.MIN_VALUE/2);
}
Arrays.sort(mnn,1,n+1);
for(int i=1;i<=n;i++) {
for(int j=k;j>0;j--) {
dp[i][j]=Math.max(dp[i-1][j]+mnn[i].b*(k-1), dp[i-1][j-1]+mnn[i].a+mnn[i].b*(j-1));
poss[i][j]=(dp[i][j]^dp[i-1][j])==0?false:true;
}
}
for(int i=n,j=k;i>0;i--) {
if(poss[i][j]) {
inc[i]=true;
j--;
}
}
out.println(k+2*(n-k));
for(int i=1,j=0;i<=n;i++) {
if(j==k-1) {
for(int z=1;z<=n;z++) {
if(!inc[z]) {
out.print(mnn[z].c+" "+(-mnn[z].c)+" ");
}
}
j++;
}
if(inc[i]) {
out.print(mnn[i].c+" ");
j++;
}
}
out.println();
}
out.close();
}
static class pair implements Comparable<pair>{
int a,b,c;
public pair(int x,int y,int z) {
a=x;b=y;c=z;
}
public int compareTo(pair p) {
return b-p.b;
//if(a>p.a) return 1;
}
}
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
#define int long long
main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout << setprecision(15) << fixed;
int t;
cin >> t;
while(t--){
int n, k;
cin >> n >> k;
vector < vector <int> > a;//(n, vector <int> (3) );
for(int i = 0; i < n; ++i){
int c, d;
cin >> c >> d;
a.push_back({d,c,i+1});
}
sort(a.begin(), a.end());
int inf = -1e18;
vector <vector <int> > dp(n+1, vector <int> (n+1, -inf));
dp[0][0] = 0;
for(int i = 1; i <= n; ++i){
dp[i][0] = dp[i-1][0] + k*a[i-1][0];
for(int j = 1; j <= min(k, i); ++j){
dp[i][j] = max(dp[i-1][j] + (k-1)*a[i-1][0], dp[i-1][j-1] + a[i-1][1] + (j-1)*a[i-1][0]);
}
}
vector <int> b(n);
{
int r = n, e = k;
while(e > 0 && r > 0){//r > 0
if (e == r){
for(int i = r-1; i >= 0; --i){
b[i] = 1;
}
break;
}
if(dp[r-1][e] + (k-1)*a[r-1][0] <= dp[r-1][e-1] + a[r-1][1] + (e-1)*a[r-1][0]){
b[r-1] = 1;
--e;
}
--r;
}
}
cout << 2*n - k << '\n';
int r = 0;
for(int i = 0; i < n; ++i){
if(r == k-1)break;
if(b[i]) {
cout << a[i][2] << ' ';
++r;
if(r == k-1)break;
}
}
for(int i = 0; i < n; ++i){
if(!b[i]) {
cout << a[i][2] << ' ';
cout << -a[i][2] << ' ';
}
}
for(int i = n-1; i >= 0; --i){
if(b[i]) {
cout << a[i][2] << '\n';
break;
}
}
}
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
python3
|
import sys
readline = sys.stdin.readline
read = sys.stdin.read
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))
prn = lambda x: print(*x, sep='\n')
def solve():
n, k = nm()
mini = [tuple(nm()) for _ in range(n)]
mini.sort(key = lambda x: -x[1])
dp = [[-10**9]*(k+1) for _ in range(n+1)]
dp[0][0] = 0
f = [[0]*(k+1) for _ in range(n+1)]
for i in range(n):
for j in range(k):
if dp[i+1][j] < dp[i][j] + (k - 1) * mini[i][1]:
dp[i+1][j] = dp[i][j] + (k - 1) * mini[i][1]
f[i+1][j] = 1
if dp[i+1][j+1] < dp[i][j] + mini[i][0] + (k - j - 1)*mini[i][1]:
dp[i+1][j+1] = dp[i][j] + mini[i][0] + (k - j - 1)*mini[i][1]
f[i+1][j+1] = 2
cx = k
a = list()
b = list()
for i in range(n, 0, -1):
if f[i][cx] == 2:
a.append(i)
cx -= 1
else:
b.append(i)
com = list()
for x in a[:-1]:
com.append(x)
for x in reversed(b):
com.append(x)
com.append(-x)
com.append(a[-1])
print(len(com))
print(*com)
return
# solve()
T = ni()
for _ in range(T):
solve()
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
string print_iterable(T1 begin_iter, T2 end_iter, int counter) {
bool done_something = false;
stringstream res;
res << "[";
for (; begin_iter != end_iter and counter; ++begin_iter) {
done_something = true;
counter--;
res << *begin_iter << ", ";
}
string str = res.str();
if (done_something) {
str.pop_back();
str.pop_back();
}
str += "]";
return str;
}
vector<int> SortIndex(int size, std::function<bool(int, int)> compare) {
vector<int> ord(size);
for (int i = 0; i < size; i++) ord[i] = i;
sort(ord.begin(), ord.end(), compare);
return ord;
}
template <typename T>
bool MinPlace(T& a, const T& b) {
if (a > b) {
a = b;
return true;
}
return false;
}
template <typename T>
bool MaxPlace(T& a, const T& b) {
if (a < b) {
a = b;
return true;
}
return false;
}
template <typename S, typename T>
ostream& operator<<(ostream& out, const pair<S, T>& p) {
out << "{" << p.first << ", " << p.second << "}";
return out;
}
template <typename T>
ostream& operator<<(ostream& out, const vector<T>& v) {
out << "[";
for (int i = 0; i < (int)v.size(); i++) {
out << v[i];
if (i != (int)v.size() - 1) out << ", ";
}
out << "]";
return out;
}
template <class TH>
void _dbg(const char* name, TH val) {
clog << name << ": " << val << endl;
}
template <class TH, class... TA>
void _dbg(const char* names, TH curr_val, TA... vals) {
while (*names != ',') clog << *names++;
clog << ": " << curr_val << ", ";
_dbg(names + 1, vals...);
}
void solve() {
int N, K;
cin >> N >> K;
vector<int> a(N);
vector<int> b(N);
for (int i = 0; i < N; i++) cin >> a[i] >> b[i];
vector<int> ord = SortIndex(N, [&](int i, int j) { return b[i] < b[j]; });
vector<vector<int>> din(N + 1, vector<int>(N + 1, -1e9));
din[0][0] = 0;
for (int i = 0; i < N; i++) {
int it = ord[i];
for (int j = 1; j <= N; j++) {
din[i + 1][j] = max(din[i][j] + (K - 1) * b[it],
din[i][j - 1] + (j - 1) * b[it] + a[it]);
}
}
vector<bool> destroyed(N + 1, false);
int cnt = K;
for (int i = N - 1; i >= 0; i--) {
int it = ord[i];
if (din[i + 1][cnt] == din[i][cnt] + (K - 1) * b[it]) {
destroyed[i] = true;
} else
cnt--;
}
assert(cnt == 0);
vector<int> ans;
int last;
for (int i = 0; i < N; i++) {
if (!destroyed[i]) {
if (((int)((ans).size())) < K - 1)
ans.push_back(ord[i] + 1);
else
last = ord[i];
}
}
for (int i = 0; i < N; i++) {
if (destroyed[i]) {
ans.push_back(ord[i] + 1);
ans.push_back(-ord[i] - 1);
}
}
ans.push_back(last + 1);
cout << ans.size() << "\n";
for (int x : ans) cout << x << " ";
cout << "\n";
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int T;
cin >> T;
for (int t = 0; t < T; t++) solve();
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
class bipartiteGraph {
public:
int n;
vector<vector<int>> cost;
vector<int> mx, my, lx, ly, p;
vector<bool> vx, vy;
vector<int> dy, pdy;
queue<int> qu{};
int MAX{numeric_limits<int>::max()};
bipartiteGraph(int n)
: n(n),
cost(n, vector<int>(n)),
mx(n, -1),
my(n, -1),
lx(n),
ly(n),
p(n),
vx(n),
vy(n),
dy(n),
pdy(n) {}
bool DFS(int x) {
vx[x] = true;
for (int y{0}; y < n; ++y)
if (!vy[y] && lx[x] + ly[y] == cost[x][y]) {
vy[y] = true;
if (my[y] == -1 || DFS(my[y])) {
mx[x] = y, my[y] = x;
return true;
}
}
return false;
}
int Hungarian() {
for (int i{0}; i < n; ++i) {
lx[i] = ly[i] = 0;
for (int j{0}; j < n; ++j)
if (cost[i][j] != MAX) lx[i] = max(lx[i], cost[i][j]);
mx[i] = my[i] = -1;
}
for (int x{0}; x < n; ++x) {
while (true) {
fill(vx.begin(), vx.end(), false);
fill(vy.begin(), vy.end(), false);
if (DFS(x)) break;
int d{MAX};
for (int x{0}; x < n; ++x)
if (vx[x])
for (int y{0}; y < n; ++y)
if (!vy[y] && cost[x][y] != MAX)
d = min(d, lx[x] + ly[y] - cost[x][y]);
if (d == MAX) return -MAX;
for (int x{0}; x < n; ++x)
if (vx[x]) lx[x] -= d;
for (int y{0}; y < n; ++y)
if (vy[y]) ly[y] += d;
}
}
for (int x{0}; x < n; ++x)
for (int y{0}; y < n; ++y)
if (lx[x] + ly[y] < cost[x][y]) {
cout << "Err1!! " << lx[x] << ' ' << ly[y] << ' ' << cost[x][y]
<< endl;
}
for (int x{0}; x < n; ++x)
if (lx[x] + ly[mx[x]] != cost[x][mx[x]]) cout << "Err2!!" << endl;
for (int y{0}; y < n; ++y)
if (lx[my[y]] + ly[y] != cost[my[y]][y]) cout << "Err3!!" << endl;
int weight{0};
for (int x{0}; x < n; ++x) weight += cost[x][mx[x]];
return weight;
}
};
void solve() {
int n, k;
cin >> n >> k;
bipartiteGraph bg(n);
for (int i{0}; i < n; ++i) {
int a, b;
cin >> a >> b;
for (int j{0}; j < n; ++j)
bg.cost[i][j] = a * (j < k - 1 || j == n - 1) + b * min(j, k - 1);
}
cout << bg.Hungarian() << '\n';
bg.Hungarian();
cout << n + (n - k) << '\n';
for (int y{0}; y < n; ++y) {
cout << bg.my[y] + 1 << ' ';
if (y >= k - 1 && y != n - 1) cout << -1 * (bg.my[y] + 1) << ' ';
}
cout << endl;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int t;
cin >> t;
while (t--) solve();
return 0;
int n;
cin >> n;
bipartiteGraph bg(n);
for (int i{0}; i < n; ++i)
for (int j{0}; j < n; ++j) cin >> bg.cost[i][j], bg.cost[i][j] *= -1;
cout << bg.Hungarian() << endl;
for (auto& y : bg.mx) cout << y << ' ';
cout << endl;
return 0;
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
long long inf = 1LL << 50;
const int N = 75;
int main() {
ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
cin.exceptions(cin.failbit);
int T;
cin >> T;
for (int z = 0; z < T; z++) {
int n, k;
cin >> n >> k;
vector<vector<pair<long long, vector<int>>>> dp(
n, vector<pair<long long, vector<int>>>(k + 1, {-inf, vector<int>()}));
vector<pair<int, pair<int, int>>> srted(n);
for (int i = 0; i < n; i++) {
int a, b;
cin >> a >> b;
srted[i].first = b;
srted[i].second.first = a;
srted[i].second.second = i;
}
sort(srted.begin(), srted.end());
dp[0][1].first = srted[0].second.first;
dp[0][1].second.push_back(srted[0].second.second);
dp[0][0].first = (n - 1) * srted[0].first;
for (int i = 1; i < n; i++) {
dp[i][0].first = dp[i - 1][0].first + srted[i].first * (n - 1);
for (int a = 1; a <= k; a++) {
if (dp[i - 1][a].first + srted[i].first * (n - 1) >
dp[i - 1][a - 1].first + srted[i].first * (a - 1) +
srted[i].second.first) {
dp[i][a].first = dp[i - 1][a].first + srted[i].first * (n - 1);
dp[i][a].second = dp[i - 1][a].second;
} else {
dp[i][a].first = dp[i - 1][a - 1].first + srted[i].first * (a - 1) +
srted[i].second.first;
dp[i][a].second = dp[i - 1][a - 1].second;
dp[i][a].second.push_back(srted[i].second.second);
}
}
}
cout << n + (n - k) << "\n";
vector<bool> visited(n, false);
for (int i = 0; i < dp[n - 1][k].second.size() - 1; i++) {
cout << dp[n - 1][k].second[i] + 1 << " ";
visited[dp[n - 1][k].second[i]] = true;
}
for (int i = 0; i < n; i++) {
if (!visited[i] && i != dp[n - 1][k].second.back()) {
cout << i + 1 << " " << -(i + 1) << " ";
}
}
cout << dp[n - 1][k].second.back() + 1 << '\n';
}
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int N = 77;
int T, n, k, f[N][N];
bool g[N][N], c[N];
void qr(int &x) { scanf("%d", &x); }
struct node {
int a, b, id;
bool operator<(node t) const { return b < t.b; }
} a[N];
int main() {
qr(T);
while (T--) {
qr(n);
qr(k);
for (int i = 1; i <= n; i++) qr(a[i].a), qr(a[i].b), a[i].id = i;
sort(a + 1, a + n + 1);
memset(f, 0xcf, sizeof f);
for (int i = 0; i <= n; i++) f[i][0] = 0;
memset(c, 0, sizeof c);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= k; j++)
f[i][j] = max(f[i - 1][j], f[i - 1][j - 1] + a[i].a +
(j == k ? 0 : (j - 1) * a[i].b)),
g[i][j] = f[i][j] ^ f[i - 1][j];
for (int i = n, j = k; j; i--)
if (g[i][j]) c[i] = j--;
printf("%d\n", k + (n - k) * 2);
for (int i = 1, j = 0; i <= n; i++) {
if (j == k - 1) {
for (int x = 1; x <= n; x++)
if (!c[x]) printf("%d %d ", a[x].id, -a[x].id);
j++;
}
if (c[i]) printf("%d ", a[i].id), j++;
}
puts("");
}
return 0;
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 β€ T β€ 75) β the number of test cases.
Each test case begins with a line containing two integers n and k (1 β€ k β€ n β€ 75) β the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 β€ a_i β€ 10^5, 0 β€ b_i β€ 10^5) β the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m β the number of actions which Polycarp has to perform (0 β€ m β€ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
template <class T>
inline void amin(T &x, const T &y) {
if (y < x) x = y;
}
template <class T>
inline void amax(T &x, const T &y) {
if (x < y) x = y;
}
template <class Iter>
void rprintf(const char *fmt, Iter begin, Iter end) {
for (bool sp = 0; begin != end; ++begin) {
if (sp)
putchar(' ');
else
sp = true;
printf(fmt, *begin);
}
putchar('\n');
}
struct Data {
long long a, b;
int i;
bool operator<(const Data &o) const { return b < o.b; }
};
int N, K;
Data P[77];
long long dp[77][77];
long long solve(int last) {
P[N] = P[last];
memset(dp, 0xc0, sizeof dp);
dp[0][0] = 0;
for (int i = 0, i_len = (N + 1); i < i_len; ++i) {
for (int j = 0, j_len = (K); j < j_len; ++j) amax(dp[i + 1][j], dp[i][j]);
if (i == last) continue;
for (int j = 0, j_len = (K); j < j_len; ++j)
amax(dp[i + 1][j + 1], dp[i][j] + P[i].a + j * P[i].b);
for (int j = 0, j_len = (K); j < j_len; ++j)
amax(dp[i + 1][j], dp[i][j] + (K - 1) * P[i].b);
}
return dp[N + 1][K];
}
void MAIN() {
scanf("%d%d", &N, &K);
for (int i = 0, i_len = (N); i < i_len; ++i) {
scanf("%lld%lld", &P[i].a, &P[i].b);
P[i].i = i;
}
sort(P, P + N);
pair<long long, int> best(-1, 0);
for (int i = 0, i_len = (N); i < i_len; ++i) {
long long tmp = solve(i);
amax(best, make_pair(tmp, i));
}
solve(best.second);
int x = K;
vector<int> ans;
for (int i = N + 1; i--;) {
if (i == best.second) continue;
if (x < K && dp[i + 1][x] == dp[i][x]) {
} else if (dp[i + 1][x] == dp[i][x - 1] + P[i].a + (x - 1) * P[i].b) {
ans.push_back(P[i].i + 1);
x--;
} else if (dp[i + 1][x] == dp[i][x] + (K - 1) * P[i].b) {
ans.push_back(-(P[i].i + 1));
ans.push_back(P[i].i + 1);
} else {
assert(false);
}
}
assert(x == 0);
printf("%d\n", (int)ans.size());
rprintf("%d", ans.rbegin(), ans.rend());
}
int main() {
int TC = 1;
scanf("%d", &TC);
for (int tc = 0, tc_len = (TC); tc < tc_len; ++tc) MAIN();
return 0;
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
template <typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;
int n, k, a, b;
long long t;
vector<long long> ve[2][2];
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cin >> n >> k;
for (int i = (0); i < (n); i++) {
cin >> t >> a >> b;
ve[a][b].push_back(t);
}
sort((ve[0][1]).begin(), (ve[0][1]).end());
sort((ve[1][0]).begin(), (ve[1][0]).end());
vector<long long> sums;
for (long long x : ve[1][1]) sums.push_back(x);
for (int i = 0; i < min(((int)(ve[0][1]).size()), ((int)(ve[1][0]).size()));
i++) {
sums.push_back(ve[0][1][i] + ve[1][0][i]);
}
sort((sums).begin(), (sums).end());
if (((int)(sums).size()) < k) {
cout << "-1\n";
return 0;
}
cout << accumulate(sums.begin(), sums.begin() + k, 0LL) << '\n';
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
import sys
input=sys.stdin.readline
f=lambda :list(map(int, input().strip('\n').split()))
n, k=f()
_11=[]
_01=[]
_10=[]
for _ in range(n):
t, a, b=f()
if a and b:
_11.append(t)
elif a:
_10.append(t)
elif b:
_01.append(t)
_01.sort(); _10.sort(); _11.sort()
for i in range(1, len(_01)):
_01[i]+=_01[i-1]
for i in range(1, len(_10)):
_10[i]+=_10[i-1]
for i in range(1, len(_11)):
_11[i]+=_11[i-1]
ans=3*1e9
if len(_01)>=k and len(_10)>=k:
ans=min(ans, _01[k-1]+_10[k-1])
for i in range(len(_11)):
if i+1<k and (len(_01)>=k-i-1) and (len(_10)>=k-i-1):
ans=min(ans, _11[i]+_01[k-i-2]+_10[k-i-2])
else:
if len(_11)>=k:
ans=min(ans, _11[k-1])
break
print(-1 if ans==3*1e9 else ans)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
const long long INFLL = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n, k;
cin >> n >> k;
vector<int> alice, bobo, osdois;
for (int i = 0; i < n; i++) {
bool a, b;
int t;
cin >> t >> a >> b;
if (a and b) {
osdois.push_back(t);
} else if (a) {
alice.push_back(t);
} else if (b) {
bobo.push_back(t);
}
}
sort(alice.begin(), alice.end());
sort(bobo.begin(), bobo.end());
sort(osdois.begin(), osdois.end());
queue<int> al, bb, ab;
for (auto num : alice) al.push(num);
for (auto num : bobo) bb.push(num);
for (auto num : osdois) ab.push(num);
int ans = 0;
int cont = 0;
while (cont < k) {
if (ab.empty()) {
if (bb.empty() or al.empty()) break;
ans += bb.front();
ans += al.front();
bb.pop();
al.pop();
cont++;
continue;
}
if (al.empty() or bb.empty()) {
ans += ab.front();
cont++;
ab.pop();
} else {
if (al.front() + bb.front() < ab.front()) {
ans += al.front() + bb.front();
al.pop();
bb.pop();
cont++;
} else {
ans += ab.front();
cont++;
ab.pop();
}
}
}
if (cont != k) {
cout << -1 << endl;
return 0;
}
cout << ans << endl;
return 0;
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, k, res = 0;
cin >> n >> k;
vector<int> arrAB;
vector<int> arrA;
vector<int> arrB;
for (int i = 0; i < n; i++) {
int time, a, b;
cin >> time >> a >> b;
if (a == 1 && b == 1) {
arrAB.push_back(time);
} else if (a == 1) {
arrA.push_back(time);
} else if (b == 1) {
arrB.push_back(time);
}
}
sort(arrA.begin(), arrA.end());
sort(arrB.begin(), arrB.end());
for (int i = 0; i < arrA.size() && i < arrB.size(); i++) {
arrAB.push_back(arrA[i] + arrB[i]);
}
sort(arrAB.begin(), arrAB.end());
if (arrAB.size() < k)
cout << "-1" << endl;
else {
for (int i = 0; i < k; i++) res += arrAB[i];
cout << res << endl;
}
return 0;
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class Codeforces
{
public static void main(String args[])throws Exception
{
BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb=new StringBuilder();
String s[]=bu.readLine().split(" ");
int n=Integer.parseInt(s[0]),k=Integer.parseInt(s[1]);
ArrayList<Integer> ab=new ArrayList<>();
ArrayList<Integer> a=new ArrayList<>();
ArrayList<Integer> b=new ArrayList<>();
int i,al=0,bo=0,x,y,z;
for(i=0;i<n;i++)
{
s=bu.readLine().split(" ");
x=Integer.parseInt(s[0]); y=Integer.parseInt(s[1]); z=Integer.parseInt(s[2]);
if(y==1) al++;
if(z==1) bo++;
if(y==1 && z==1) ab.add(x);
else if(y==1) a.add(x);
else if(z==1) b.add(x);
}
if(al<k || bo<k) {System.out.print("-1"); return;}
Collections.sort(ab); Collections.sort(a); Collections.sort(b);
ArrayList<Integer> alb=new ArrayList<>();
for(i=0;i<Math.min(a.size(),b.size());i++)
alb.add(a.get(i)+b.get(i));
int min=0,c=0;
if(alb.size()==0)
{
for(i=0;i<k;i++)
min+=ab.get(i);
System.out.print(min);
return;
}
if(ab.size()==0)
{
for(i=0;i<k;i++)
min+=alb.get(i);
System.out.print(min);
return;
}
x=0; y=0;
while(x<ab.size() && y<alb.size() && c<k)
{
if(ab.get(x)<alb.get(y)) {min+=ab.get(x); x++;}
else {min+=alb.get(y); y++;}
c++;
}
if(c==k) {System.out.print(min); return;}
while(x<ab.size() && c<k)
{
min+=ab.get(x);
x++;
c++;
}
while(y<alb.size() && c<k)
{
min+=alb.get(y);
y++;
c++;
}
System.out.print(min);
}
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
from collections import deque
n, k = map(int, input().split())
onla = []
onlb = []
both = []
for i in range(n):
a, b, c = map(int, input().split())
if b == 1 and c == 1:
both.append(a)
elif c == 1:
onlb.append(a)
elif b == 1:
onla.append(a)
if len(both) + len(onla) < k or len(both) + len(onlb) < k:
print(-1)
else:
onlb.sort()
onlb = deque(onlb)
onla.sort()
onla = deque(onla)
both.sort()
both = deque(both)
al = k
bob = k
ans = 0
while al > 0 or bob > 0:
if al > 0 and bob > 0:
if len(onla) == 0 or len(onlb) == 0:
ans += both[0]
al -= 1
bob -= 1
both.popleft()
else:
if len(both) == 0:
ans += onla[0] + onlb[0]
al -= 1
bob -= 1
onla.popleft()
onlb.popleft()
else:
if onla[0] + onlb[0] <= both[0]:
ans += onla[0] + onlb[0]
al -= 1
bob -= 1
onla.popleft()
onlb.popleft()
else:
ans += both[0]
al -= 1
bob -= 1
both.popleft()
elif bob > 0:
if both[0] < onlb[0]:
ans += both[0]
bob -= 1
both.popleft()
else:
ans += onlb[0]
bob -= 1
onlb.popleft()
elif al > 0:
if both[0] < onla[0]:
ans += both[0]
bob -= 1
both.popleft()
else:
ans += onla[0]
bob -= 1
onlb.popleft()
print(ans)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
class Book:
def __init__(self, time, for_alice, for_bob):
self.time = time
self.for_alice = bool(for_alice)
self.for_bob = bool(for_bob)
self.status = ('A' if self.for_alice else '') + ('B' if self.for_bob else '')
def get(li, i):
try:
return li[i].time
except IndexError:
return 10 ** 10
n, k = map(int, input().split())
books = [Book(*map(int, input().split())) for _ in range(n)]
alice_rest, bob_rest = k, k
for_alice_books = list(filter(lambda x: x.status == 'A', books))
for_bob_books = list(filter(lambda x: x.status == 'B', books))
for_both_books = list(filter(lambda x: x.status == 'AB', books))
for_alice_books = sorted(for_alice_books, key=lambda x: x.time)
for_bob_books = sorted(for_bob_books, key=lambda x: x.time)
for_both_books = sorted(for_both_books, key=lambda x: x.time)
for_alice_books_index, for_bob_books_index, for_both_books_index = 0, 0, 0
total = 0
while alice_rest > 0 or bob_rest > 0:
alice_var = get(for_alice_books, for_alice_books_index)
bob_var = get(for_bob_books, for_bob_books_index)
both_var = get(for_both_books, for_both_books_index)
if all(i == 10 ** 10 for i in [alice_var, bob_var, both_var]):
total = -1
break
if alice_rest > 0 and bob_rest > 0:
if alice_var + bob_var > both_var:
for_both_books_index += 1
total += both_var
else:
for_alice_books_index += 1
for_bob_books_index += 1
total += alice_var + bob_var
alice_rest -= 1
bob_rest -= 1
elif alice_rest == 0:
if bob_var > both_var:
for_both_books_index += 1
total += both_var
else:
for_bob_books_index += 1
total += bob_var
bob_rest -= 1
else:
if alice_var > both_var:
for_both_books_index += 1
total += both_var
else:
for_alice_books_index += 1
total += alice_var
alice_rest -= 1
print(total if total < 10 ** 10 else -1)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
from sys import stdin
input=stdin.readline
def answer():
if(n3+n1 < k or n3+n2 < k):return -1
i,j=0,0
ans=0
for take in range(k):
if(i >= n1 or i >= n2):
ans+=common[j]
j+=1
elif(j >= n3):
ans+=a[i]+b[i]
i+=1
else:
if(a[i]+b[i] > common[j]):
ans+=common[j]
j+=1
else:
ans+=a[i]+b[i]
i+=1
return ans
n,k=map(int,input().split())
a,b,common=[],[],[]
for i in range(n):
t,x,y=map(int,input().split())
if(x and y):common.append(t)
elif(x==1 and y==0):a.append(t)
elif(x==0 and y==1):b.append(t)
common.sort()
a.sort()
b.sort()
n1,n2,n3=len(a),len(b),len(common)
print(answer())
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
nb_books, mins = [int(x) for x in input().split()]
x, y, z = [],[],[]
for i in range(nb_books):
n, a, b = [int(x) for x in input().split()]
if a & b:z.append(n)
elif a:x.append(n)
elif b:y.append(n)
x.sort()
y.sort()
for i in range(min(len(x), len(y))) :
z.append(x[i] + y[i])
answer = sum(sorted(z)[:mins])
print(-1 if len(z) < mins else answer)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
import java.util.*;
public class readingBooks {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int n = input.nextInt();
int k = input.nextInt();
ArrayList<Integer>[] arr = new ArrayList[4];
for(int i = 0; i < 4; i++){
arr[i] = new ArrayList<Integer>();
}
for(int i = 0; i < n; i++){
int p = input.nextInt();
int bin = input.nextInt() * 2 + input.nextInt();
arr[bin].add(p);
}
if(arr[3].size() + Math.min(arr[1].size(), arr[2].size()) < k){
System.out.println(-1);
return;
}
for(int i = 0; i < 4; i++){
Collections.sort(arr[i]);
}
int a, b, ans;
a = b = ans = 0;
for(int i = 0; i < k; i ++){
boolean A = a < arr[3].size(), B = b >= arr[1].size() || b >= arr[2].size();
if(B || (A && arr[3].get(a) < arr[1].get(b) + arr[2].get(b))){
ans += arr[3].get(a);
a ++;
}else{
ans += arr[1].get(b) + arr[2].get(b);
b ++;
}
}
System.out.println(ans);
}
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
import java.math.*;
import java.lang.*;
import static java.lang.Math.*;
public class Solution implements Runnable {
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
private BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public double nextDouble() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E')
return res * Math.pow(10, nextInt());
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E')
return res * Math.pow(10, nextInt());
if (c < '0' || c > '9')
throw new InputMismatchException();
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return readString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
public static void main(String args[]) throws Exception {
new Thread(null, new Solution(), "Main", 1 << 27).start();
}
static class Pair {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + x * 7 + (y * 3 + 5 * (y - x));
return result;
}
@Override
public boolean equals(Object obj) {
if (this == obj) {
return true;
}
if (obj == null) {
return false;
}
if (getClass() != obj.getClass()) {
return false;
}
Pair other = (Pair) obj;
if (x != other.x && y != other.y) {
return false;
}
return true;
}
}
static void sieveOfEratosthenes(int n) {
//Prints prime nos till n
boolean prime[] = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++) {
if (prime[p] == true) {
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
for (int i = 2; i <= n; i++) {
if (prime[i] == true)
System.out.print(i + " ");
}
}
public void run() {
InputReader in = new InputReader(System.in);
PrintWriter w = new PrintWriter(System.out);
int n=in.nextInt();
int k=in.nextInt();
ArrayList<Integer> A=new ArrayList<Integer>();
ArrayList<Integer> B=new ArrayList<Integer>();
ArrayList<Integer> AB=new ArrayList<Integer>();
for(int i=0;i<n;i++)
{
int t=in.nextInt();
int a=in.nextInt();
int b=in.nextInt();
if(a==1 && b==1)
AB.add(t);
else if(a==1 && b==0)
A.add(t);
else if(a==0 && b==1)
B.add(t);
}
Collections.sort(A);
Collections.sort(B);
Collections.sort(AB);
if((A.size()+AB.size())<k)
w.println(-1);
else if((B.size()+AB.size())<k)
w.println(-1);
else
{
long count=0;
if(A.size()==0 || B.size()==0)
{
for(int i=0;i<AB.size() && i<k;i++)
count+=AB.get(i);
w.println(count);
}
else
{
int i=0,j=0,x=0;
while(i<A.size() && i<B.size() && j<AB.size() && x<k)
{
if((A.get(i)+B.get(i))<(AB.get(j)))
{
count+=A.get(i)+B.get(i);
i++;
}
else
{
count+=AB.get(j);
j++;
}
x++;
}
while(x<k && j<AB.size())
{
count+=AB.get(j);
j++;
x++;
}
while(x<k && i<A.size() && i<B.size())
{
count+=A.get(i)+B.get(i);
i++;
x++;
}
w.println(count);
}
}
w.flush();
w.close();
}
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
public final class ReadingBooksEasy {
private static final FastReader fr = new FastReader();
public static void main(String[] args) {
final int n = fr.nextInt(), k = fr.nextInt();
final Triple[] triples = new Triple[n];
for (int i = 0; i < n; i++) {
triples[i] = new Triple(fr.nextInt(), fr.nextInt() == 1, fr.nextInt() == 1);
}
int ans = new ReadingBooksEasy(triples, k).solve();
System.out.println(ans);
}
private final Triple[] triples;
private final int k;
public ReadingBooksEasy(final Triple[] triples, final int k) {
this.triples = triples;
this.k = k;
}
private int solve() {
int[] bothCumulative, aliceCumulative, bobCumulative;
{
final List<Integer> both = new LinkedList<>(), alice = new LinkedList<>(), bob = new LinkedList<>();
for (Triple triple : triples) {
if (triple.alice && triple.bob) both.add(triple.time);
else if (triple.alice) alice.add(triple.time);
else if (triple.bob) bob.add(triple.time);
}
final int[] bothArr = both.stream().mapToInt(Integer::intValue).toArray(),
aliceArr = alice.stream().mapToInt(Integer::intValue).toArray(),
bobArr = bob.stream().mapToInt(Integer::intValue).toArray();
Arrays.sort(bothArr);
Arrays.sort(aliceArr);
Arrays.sort(bobArr);
bothCumulative = new int[both.size() + 1];
aliceCumulative = new int[alice.size() + 1];
bobCumulative = new int[bob.size() + 1];
System.arraycopy(bothArr, 0, bothCumulative, 1, bothArr.length);
System.arraycopy(aliceArr, 0, aliceCumulative, 1, aliceArr.length);
System.arraycopy(bobArr, 0, bobCumulative, 1, bobArr.length);
cumulative(bothCumulative);
cumulative(aliceCumulative);
cumulative(bobCumulative);
}
final List<Integer> answers = new ArrayList<>(k);
for (int common = k; common >= 0; common--) {
if (bothCumulative.length - 1 < common || aliceCumulative.length - 1 < k - common || bobCumulative.length - 1 < k - common)
continue;
int time = bothCumulative[common]
+ aliceCumulative[k - common]
+ bobCumulative[k - common];
answers.add(time);
}
return answers.stream().min(Integer::compare).orElse(-1);
}
private static void cumulative(final int[] arr) {
if (arr.length == 0) return;
for (int i = 1; i < arr.length; i++) {
arr[i] = arr[i - 1] + arr[i];
}
}
private static final class Triple {
public final int time;
public final boolean alice, bob;
public Triple(int time, boolean alice, boolean bob) {
this.time = time;
this.alice = alice;
this.bob = bob;
}
}
private static final class FastReader {
private final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
private StringTokenizer st;
public String nextLine() {
try {
return br.readLine();
} catch (IOException ex) {
throw new RuntimeException(ex);
}
}
public String next() {
while (st == null || !st.hasMoreTokens()) {
st = new StringTokenizer(nextLine());
}
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
n, k = map(int, input().split())
books = []
alice = []
bob = []
for _ in range(n):
book = tuple(map(int, input().split()))
if book[1] == book[2] == 1:
books.append(book[0])
elif book[1] == 1:
alice.append(book[0])
elif book[2] == 1:
bob.append(book[0])
for a, b in zip(sorted(alice), sorted(bob)):
books.append(a+b)
print(sum(sorted(books)[:k]) if len(books) >= k else -1)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;
public class cdf653e1
{
static void merge(int arr[], int l, int m, int r)
{
int n1 = m - l + 1;
int n2 = r - m;
int L[] = new int [n1];
int R[] = new int [n2];
for (int i=0; i<n1; ++i)
L[i] = arr[l + i];
for (int j=0; j<n2; ++j)
R[j] = arr[m + 1+ j];
int i = 0, j = 0;
int k = l;
while (i < n1 && j < n2)
{
if (L[i] <= R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
static void sort(int arr[], int l, int r)
{
if (l < r)
{
int m = (l+r)/2;
sort(arr, l, m);
sort(arr , m+1, r);
merge(arr, l, m, r);
}
}
public static int lowerBound(int[] array, int length, int value)
{
int low = 0;
int high = length;
while (low < high)
{
final int mid = (low + high) / 2;
//checks if the value is less than middle element of the array
if (value <= array[mid])
{
high = mid;
}
else
{
low = mid + 1;
}
}
return low;
}
public static int upperBound(int[] array, int length, int value)
{
int low = 0;
int high = length;
while (low < high) {
final int mid = (low + high) / 2;
if (value >= array[mid]) {
low = mid + 1;
} else {
high = mid;
}
}
return low;
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static long power(long n,long m)
{
if(m==0)
return 1;
long ans=1;
while(m>0)
{
ans=ans*n;
m--;
}
return ans;
}
static int BinarySearch(int arr[], int x)
{
int l = 0, r = arr.length - 1;
while (l <= r) {
int m = l + (r - l) / 2;
if (arr[m] == x)
return m;
if (arr[m] < x)
l = m + 1;
else
r = m - 1;
}
return -1;
}
public static void main(String args[])throws Exception
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int t=1;//Integer.parseInt(br.readLine());
for(int i=1;i<=t;i++)
{
String s=br.readLine();
String str[]=s.split(" ");
int n=Integer.parseInt(str[0]);
int k=Integer.parseInt(str[1]);
ArrayList <Integer> ab=new ArrayList<Integer>();
ArrayList <Integer> a=new ArrayList<Integer>();
ArrayList <Integer> b=new ArrayList<Integer>();
for(int j=0;j<n;j++)
{
s=br.readLine();
String str1[]=s.split(" ");
int temp=Integer.parseInt(str1[0]);
int temp1=Integer.parseInt(str1[1]);
int temp2=Integer.parseInt(str1[2]);
if(temp1==1&&temp2==1)
{
ab.add(temp);
continue;
}
else if(temp1==1)
{
a.add(temp);
continue;
}
else if(temp2==1)
{
b.add(temp);
continue;
}
}
Collections.sort(ab);
Collections.sort(a);
Collections.sort(b);
int ak=k,bk=k;
if(ab.size()+a.size()<k||ab.size()+b.size()<k)
{
System.out.println("-1");
break;
}
int pos=0;
long ans=0;
for(int j=0;j<ab.size()&&ak>0&&bk>0;)
{
if(pos<a.size()&&pos<b.size()&&ab.get(j)<=a.get(pos)+b.get(pos))
{
ans+=ab.get(j);
ak--;
bk--;
j++;
}
else if(pos<a.size()&&pos<b.size())
{
ans+=a.get(pos);
ans+=b.get(pos);
ak--;
bk--;
pos++;
}
else
{
ans+=ab.get(j);
ak--;
bk--;
j++;
}
}
int tt=pos;
while(ak>0)
{
ans+=a.get(pos);
pos++;
ak--;
}
while(bk>0)
{
ans+=b.get(tt);
tt++;
bk--;
}
System.out.println(ans);
}
}
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long maxn = 15e5 + 10;
const long long inf = 1e14;
long long n, m, k, t;
long long cur1, cur2;
long long C[maxn];
vector<long long> ANS;
vector<pair<long long, long long> > X[4];
void add(vector<long long> good) {
long long mn = inf + 1, mni = 3;
for (long long q = 0; q < 4; q++) {
if (!good[q]) {
continue;
}
if (X[q].back().first < mn) {
mn = X[q].back().first;
mni = q;
}
}
ANS.push_back(X[mni].back().second);
X[mni].pop_back();
if (mni / 2 == 1) {
cur2 = max(0LL, cur2 - 1);
}
if (mni % 2 == 1) {
cur1 = max(0LL, cur1 - 1);
}
}
void add1() { add({0, 1, 0, 1}); }
void add2() { add({0, 0, 1, 1}); }
void add0() { add({1, 1, 1, 1}); }
void add3() { add({0, 0, 0, 1}); }
void add_def() {
long long q, w, e;
vector<long long> mni = {0, 1};
long long mn = inf;
for (q = 0; q < 4; q++) {
for (w = q + 1; w < 4; w++) {
if ((q | w) != 3) {
continue;
}
vector<long long> dmni = {q, w};
long long dmn = X[q].back().first + X[w].back().first;
if (dmn < mn) {
mn = dmn;
mni = dmni;
}
}
}
for (auto i : mni) {
ANS.push_back(X[i].back().second);
X[i].pop_back();
if (i / 2 == 1) {
cur2 = max(cur2 - 1, 0LL);
}
if (i % 2 == 1) {
cur2 = max(cur2 - 1, 0LL);
}
}
}
bool used[maxn];
long long I[maxn];
int main() {
ios_base ::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
long long q, w, e, a, b, c;
cin >> n >> m >> k;
cur1 = cur2 = k;
vector<pair<long long, long long> > ALL;
for (q = 0; q < n; q++) {
cin >> a >> b >> c;
C[q] = a;
long long i = b * 2 + c;
X[i].push_back(make_pair(a, q));
ALL.push_back(make_pair(a, q));
}
for (q = 0; q < n; q++) {
for (w = 0; w < 4; w++) {
X[w].push_back(make_pair(inf, n + q * 4 + w));
ALL.push_back(make_pair(inf, n + q * 4 + w));
}
}
long long alln = n + n * 4;
sort(ALL.begin(), ALL.end());
for (q = 0; q < 4; q++) {
sort(X[q].begin(), X[q].end());
}
long long mn = 1e18, mni = 0;
long long cur = 0;
for (q = 0; q < k; q++) {
cur += X[3][q].first;
used[X[3][q].second] = 1;
}
for (q = 0; q < ALL.size(); q++) {
I[ALL[q].second] = q;
}
long long cnt = 0;
for (q = 0; cnt < m - k; q++) {
if (used[ALL[q].second]) {
continue;
}
cur += ALL[q].first;
cnt++;
}
long long iall = q - 1;
if (cur < mn) {
mn = cur;
mni = k;
}
for (q = k - 1; q >= 0; q--) {
if (q + 2 * (k - q) > m) {
break;
}
used[X[3][q].second] = 0;
cur -= X[3][q].first;
if (I[X[3][q].second] <= iall) {
cur -= ALL[iall].first;
cur += X[3][q].first;
iall--;
while (iall >= 0 && used[ALL[iall].second]) {
iall--;
}
}
long long i1 = k - q - 1;
used[X[2][i1].second] = 1;
cur += X[2][i1].first;
if (I[X[2][i1].second] <= iall) {
cur -= X[2][i1].first;
iall++;
while (used[ALL[iall].second]) {
iall++;
}
cur += ALL[iall].first;
}
used[X[1][i1].second] = 1;
cur += X[1][i1].first;
if (I[X[1][i1].second] <= iall) {
cur -= X[1][i1].first;
iall++;
while (used[ALL[iall].second]) {
iall++;
}
cur += ALL[iall].first;
}
cur -= ALL[iall].first;
iall--;
while (iall >= 0 && used[ALL[iall].second]) {
iall--;
}
if (cur < mn) {
mn = cur;
mni = q;
}
}
for (q = 0; q < alln; q++) {
used[q] = 0;
}
for (q = 0; q < mni; q++) {
ANS.push_back(X[3][q].second);
used[X[3][q].second] = 1;
}
for (q = 0; q < k - mni; q++) {
ANS.push_back(X[1][q].second);
used[X[1][q].second] = 1;
ANS.push_back(X[2][q].second);
used[X[2][q].second] = 1;
}
cnt = 0;
for (q = 0; cnt < m - mni - 2 * (k - mni); q++) {
if (!used[ALL[q].second]) {
ANS.push_back(ALL[q].second);
cnt++;
}
}
for (auto i : ANS) {
if (i >= n) {
cout << -1;
return 0;
}
}
cout << mn << endl;
for (q = 0; q < ANS.size(); q++) {
cout << ANS[q] + 1 << " ";
}
return 0;
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
from sys import stdin,stdout
# stdin = open("input.txt","r")
# stdout = open("output.txt","w")
n,k = stdin.readline().strip().split(' ')
n,k = int(n),int(k)
books=[]
for i in range(n):
t,a,b=stdin.readline().strip().split(' ')
t,a,b=int(t),int(a),int(b)
books.append((t,a,b))
pairs_a=[]
pairs_b=[]
single=[]
for i in books:
if i[1]==1 and i[2]==1:
single.append(i[0])
elif i[1] or i[2]==1:
if i[1]==1:
pairs_a.append(i[0])
else:
pairs_b.append(i[0])
pairs_b=sorted(pairs_b)
pairs_a=sorted(pairs_a)
# print(single)
# print(pairs_a)
# print(pairs_b)
pairs=[]
for i in range(min(len(pairs_a),len(pairs_b))):
single.append(pairs_a[i]+pairs_b[i])
single=sorted(single)
if len(single)<k:
stdout.write(str(-1)+"\n")
else:
stdout.write(str(sum(single[:k]))+"\n")
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
n, k = map(int, input().split())
alice=[];bob=[];both=[]
for _ in range(n):
x,y,z=map(int,input().split())
if y==1 and z==1:
both.append(x)
elif y==1:
alice.append(x)
elif z==1:
bob.append(x)
alice.sort();bob.sort()
# print(alice)
for i in range(min(len(alice),len(bob))):
both.append(alice[i]+bob[i])
both.sort()
# print(both)
if len(both)<k:
print(-1)
else:
print(sum(both[:k]))
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int solve() {
int n, k;
cin >> n >> k;
int a, b, ans;
ans = a = b = 0;
priority_queue<int> x, y, z;
while (n--) {
int t, u, v;
cin >> t >> u >> v;
t *= -1;
if (u && v) {
z.push(t);
} else if (u) {
x.push(t);
} else if (v) {
y.push(t);
}
}
while (!x.empty() && !y.empty() && !z.empty() && a < k) {
int u, v, w;
u = -x.top();
v = -y.top();
w = -z.top();
if (u + v < w) {
ans += u + v;
x.pop();
y.pop();
} else {
ans += w;
z.pop();
}
++a;
++b;
}
while (a < k && !z.empty() && x.empty()) {
ans += -z.top();
z.pop();
++a;
++b;
}
while (b < k && !z.empty() && y.empty()) {
ans += -z.top();
z.pop();
++a;
++b;
}
while (a < k && !x.empty() && !z.empty()) {
int u = -x.top();
int w = -z.top();
if (u < w) {
ans += u;
++a;
x.pop();
} else {
ans += w;
++a;
++b;
z.pop();
}
}
while (a < k && !x.empty()) {
ans += -x.top();
x.pop();
++a;
}
while (b < k && !y.empty() && !z.empty()) {
int u = -y.top();
int w = -z.top();
if (u < w) {
ans += u;
++b;
y.pop();
} else {
ans += w;
++a;
++b;
z.pop();
}
}
while (b < k && !y.empty()) {
ans += -y.top();
y.pop();
++b;
}
if (a < k || b < k)
cout << "-1\n";
else
cout << ans << '\n';
return 0;
}
int main() {
cin.tie(0)->sync_with_stdio(0);
int t = 1;
while (t--) solve();
return 0;
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
def main():
n,k=list(map(int,input().split()))
l=[]
c1=0
c2=0
for j in range(0,n):
l1=list(map(int,input().split()))
if l1[1]==1:
c1+=1
if l1[2]==1:
c2+=1
l.append(l1)
if c1<k or c2<k:
print(-1)
return
c=0
d={}
l2=[]
l3=[]
l4=[]
for j in range(0,n):
if l[j][1]==1 and l[j][2]==1:
l4.append(l[j][0])
elif l[j][1]==1:
l2.append(l[j][0])
elif l[j][2]==1:
l3.append(l[j][0])
m=10**9+7
l2.append(m)
l3.append(m)
l4.append(m)
l2.sort()
l3.sort()
l4.sort()
#print(l2,l3,l4)
p1=0
p2=0
p3=0
f=0
k1=k
k2=k
j=0
while k1!=0 or k2!=0:
if l2[p1]+l3[p2] >= l4[p3] and p3!=len(l4)-1:
c+=l4[p3]
if p3!=len(l4)-1:
p3+=1
k1-=1
k2-=1
else:
if k1!=0 and p1!=len(l2)-1:
c+=l2[p1]
if p1!=len(l2)-1:
p1+=1
k1-=1
if k2!=0 and p2!=len(l3)-1:
c+=l3[p2]
if p2!=len(l3)-1:
p2+=1
k2-=1
if k1==0 and k2==0:
break
print(c)
t=1
for i in range(0,t):
main()
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush, nsmallest
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
"""
pppppppppppppppppppp
ppppp ppppppppppppppppppp
ppppppp ppppppppppppppppppppp
pppppppp pppppppppppppppppppppp
pppppppppppppppppppppppppppppppp
pppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppp
pppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppp
pppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppp
pppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppp
pppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppp
pppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppp
pppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppp
pppppppppppppppppppppppp
pppppppppppppppppppppppppppppppp
pppppppppppppppppppppp pppppppp
ppppppppppppppppppppp ppppppp
ppppppppppppppppppp ppppp
pppppppppppppppppppp
"""
n, k = sp()
mat = []
a, b, both = [], [], []
for i in range(n):
ti, ai, bi = sp()
if ai and bi:
both.append(ti)
continue
if ai:
a.append(ti)
if bi:
b.append(ti)
if len(a) + len(both) < k or len(b) + len(both) < k:
out(-1)
exit()
a.sort(reverse=True)
b.sort(reverse=True)
both.sort(reverse=True)
answer = 0
for i in range(k):
t1 = inf if not a else a[-1]
t2 = inf if not b else b[-1]
t3 = inf if not both else both[-1]
if t3 < t1 + t2:
answer += t3
both.pop()
else:
answer += t1 + t2
a.pop()
b.pop()
out(answer)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
long long r = 0, f = 0;
char c;
while (!isdigit(c = getchar())) f |= (c == '-');
while (isdigit(c)) r = (r << 1) + (r << 3) + (c ^ 48), c = getchar();
return f ? -r : r;
}
inline long long min(long long a, long long b) { return a < b ? a : b; }
struct Book {
long long t, a, b;
bool operator<(const Book &book) const {
return t ^ book.t ? t < book.t : (a & b) > (book.a & book.b);
}
} b[202202];
long long n, k, sum, ans, cnt[2], size[2], s[2][202202];
inline void work() {
n = read(), k = read();
for (long long i = 1; i <= n; i++) {
b[i].t = read();
b[i].a = read();
b[i].b = read();
if (b[i].a | b[i].b) sum += b[i].t;
cnt[0] += b[i].a;
cnt[1] += b[i].b;
}
if (cnt[0] < k || cnt[1] < k) {
puts("-1");
return;
}
sort(b + 1, b + 1 + n);
for (long long i = n; i >= 1; i--)
if (b[i].a && !b[i].b) {
s[0][size[0] + 1] = s[0][size[0]];
size[0]++;
s[0][size[0]] += b[i].t;
}
for (long long i = size[0] + 1; i <= n; i++) s[0][i] = s[0][size[0]];
for (long long i = n; i >= 1; i--)
if (b[i].b && !b[i].a) {
s[1][size[1] + 1] = s[1][size[1]];
size[1]++;
s[1][size[1]] += b[i].t;
}
for (long long i = size[1] + 1; i <= n; i++) s[1][i] = s[1][size[1]];
ans = sum - s[0][cnt[0] - k] - s[1][cnt[1] - k];
for (long long i = n; i >= 1; i--) {
if (cnt[0] == k || cnt[1] == k) break;
if (!(b[i].a & b[i].b)) continue;
cnt[0]--, cnt[1]--;
sum -= b[i].t;
ans = min(ans, sum - s[0][cnt[0] - k] - s[1][cnt[1] - k]);
}
printf("%lld", ans);
}
signed main() {
work();
return 0;
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class hello2
{static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
static String sum (String s)
{
String s1 = "";
if(s.contains("a"))
s1+="a";
if(s.contains("e"))
s1+="e";
if(s.contains("i"))
s1+="i";
if(s.contains("o"))
s1+="o";
if(s.contains("u"))
s1+="u";
return s1;
}
public static HashMap<String, Integer> sortByValue(HashMap<String, Integer> hm)
{
// Create a list from elements of HashMap
List<Map.Entry<String, Integer> > list =
new LinkedList<Map.Entry<String, Integer> >(hm.entrySet());
// Sort the list
Collections.sort(list, new Comparator<Map.Entry<String, Integer> >() {
public int compare(Map.Entry<String, Integer> o1,
Map.Entry<String, Integer> o2)
{
return (o1.getValue()).compareTo(o2.getValue());
}
});
// put data from sorted list to hashmap
HashMap<String, Integer> temp = new LinkedHashMap<String, Integer>();
for (Map.Entry<String, Integer> aa : list) {
temp.put(aa.getKey(), aa.getValue());
}
return temp;
}
public static void main(String args[])
{
FastReader input =new FastReader();
int n = input.nextInt();
int k = input.nextInt();
int arr[][] = new int[n][3];
int arr1[][] = new int[n][3];
int arr2[][] = new int[n][3];
for(int i=0;i<n;i++)
{
arr[i][0] = input.nextInt();
arr[i][1] = input.nextInt();
arr[i][2] = input.nextInt();
if(arr[i][1]==arr[i][2] && arr[i][1]==1)
{
}
else
{
arr1[i][0] = arr[i][0];
arr1[i][1] = arr[i][1];
arr1[i][2] = arr[i][2];
arr2[i][0] = arr[i][0];
arr2[i][1] = arr[i][1];
arr2[i][2] = arr[i][2];
}
}
Arrays.sort(arr1, (a, b) -> Integer.compare(a[0], b[0]));
Arrays.sort(arr1, (a, b) -> Integer.compare(b[1], a[1]));
Arrays.sort(arr2, (a, b) -> Integer.compare(a[0], b[0]));
Arrays.sort(arr2, (a, b) -> Integer.compare(b[2], a[2]));
Arrays.sort(arr, (a, b) -> Integer.compare(a[0], b[0]));
Arrays.sort(arr, (a, b) -> Integer.compare(b[2], a[2]));
Arrays.sort(arr, (a, b) -> Integer.compare(b[1], a[1]));
// for(int i=0;i<n;i++)
// {
// System.out.println(arr2[i][0] + " " + arr2[i][1] + " " + arr2[i][2]);
// }
//
// System.out.println();
//
//
// for(int i=0;i<n;i++)
// {
// System.out.println(arr2[i][0] + " " + arr2[i][1] + " " + arr2[i][2]);
// }
//
int count =0;
int count1 =0;
int count2 =0;
long sum =0;
for(int i=0;i<n;i++)
{
if(arr[i][1]==1 && arr[i][2]==1)
count2++;
else
{
if(arr[i][1]==1)
count++;
if(arr[i][2]==1)
count1++;
}
}
// System.out.println();
//
// System.out.println(count+" " + count1 + " " + count2);
// System.out.println();
//
if(count+count2<k || count1+count2<k)
{
System.out.println(-1);
}
else
{
int i=0;
int j=0;
sum =0;
int k1 = k,k2 =k;
while(k1>0 && k2>0)
{
if((arr[i][0] > (arr1[j][0] + arr2[j][0]) && count>0 && count1>0) || count2<=0)
{
k1-=1;
k2-=1;
sum+=(arr1[j][0] + arr2[j][0]);
j++;
count--;
count1--;
}
else if(count2>0)
{
k1-=1;
k2-=1;
sum+=arr[i][0];
i++;
count2--;
}
// System.out.println(sum + " "+ count+" " + count1 + " " + count2);
}
System.out.println(sum);
}
}
//
//Arrays.sort(myArr, (a, b) -> Double.compare(a[0], b[0]));
//Arrays.sort(contests, (a, b) -> Integer.compare(b[0],a[0])); decreasing order
public static boolean range(int x,int y,int n,int m)
{
if(x<0 || x>n-1 || y<0 || y>m-1)
return false;
return true;
}
public static int even_element(int b[],int n)
{
int c[] = new int[n/2 + 1];
int l=0;
for(int i=1;i<n;i+=2)
{
c[l] = b[i];
System.out.print(c[l] + " ");
l++;
}
System.out.println();
if(c.length!=1)
{
return even_element(c,n/2);
}
else
return c[0];
}
static boolean isPrime(int n)
{
// Corner case
if (n <= 1)
return false;
// Check from 2 to n-1
for (int i = 2; i < n; i++)
if (n % i == 0)
return false;
return true;
}
}
class Pair
{
int a;
int b;
Pair(int a,int b)
{
this.a=a;
this.b=b;
}
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
n, k = list(map(int, input().split()))
a = []
b = []
ab = []
for _ in range(0, n):
t, x, y = list(map(int, input().split()))
if (x == 1 and y == 1):
ab.append(t)
elif (x == 1):
a.append(t)
elif (y == 1):
b.append(t)
# print(a, b, ab)
ab.sort(reverse=True)
b.sort(reverse=True)
a.sort(reverse=True)
ca = 0
cb = 0
t = 0
c = 0
u = 0
while (len(a) != 0 and len(b) != 0):
if (len(ab) > 0 and ab[-1] <= a[-1] + b[-1]):
t += ab.pop()
c += 1
else:
t = t + a.pop() + b.pop()
c += 1
if (c == k):
u = 1
break
if (u == 1):
print(t)
elif (c + len(ab) < k):
print(-1)
else:
t += sum(ab[-(k - c):])
print(t)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
import sys
input = sys.stdin.readline
n, k = map(int, input().split())
o, a, b = [], [], []
for i in range(n):
q, w, e = map(int, input().split())
if w == e and w:
o.append(q)
elif w and not e:
a.append(q)
elif w or e:
b.append(q)
i = j = l = 0
done = 0
ans = 0
o.sort()
a.sort()
b.sort()
while (i < len(o) or j < len(a) and l < len(b)) and done < k:
if i < len(o) and (j >= len(a) or l >= len(b) or j < len(a) and l < len(b) and o[i] <= a[j]+b[l]):
ans += o[i]
i += 1
done += 1
elif l < len(b) and j < len(a):
ans += a[j]+b[l]
j += 1
l += 1
done += 1
print(ans if done >= k else -1)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
n,k = list(map(int,input().split()))
ca,cb=0,0
x,y,z = [],[],[]
for i in range(n):
t,a,b, = list(map(int,input().split()))
ca+=a
cb+=b
if(a==1 and b==0):
x.append(t)
elif(a==0 and b==1):
y.append(t)
elif(a==1 and b==1):
z.append(t)
x.sort()
y.sort()
z.sort()
if(ca<k or cb<k):
print("-1")
else:
total = 0
arr = []
for i in range(min(len(z),k)):
total += z[i]
arr.append(z[i])
if(len(z)<k):
for i in range(len(z),k):
total+= x[0]+y[0]
del x[0]
del y[0]
while(len(x)>0 and len(y)>0 and len(arr)>0 and x[0]+y[0]<arr[-1]):
total += (x[0]+y[0]-arr[-1])
del arr[-1]
del x[0]
del y[0]
print(total)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
namespace BF {
long long solve(int N, int M, int K, const vector<int>& T, const vector<int>& A,
const vector<int>& B) {
long long res = INF;
for (int mask = 0, _n = (1 << N); mask < _n; ++mask) {
if ((__builtin_popcount((mask))) != M) continue;
int cntA = 0, cntB = 0;
long long cur = 0;
for (int i = 0, _n = (N); i < _n; ++i) {
if (mask & (1 << i)) {
if (A[i]) cntA++;
if (B[i]) cntB++;
cur += T[i];
}
}
if (cntA >= K && cntB >= K) res = min(res, cur);
}
return res < INF ? res : -1;
}
} // namespace BF
struct Item {
int id, value;
bool operator<(const Item i) const {
if (value != i.value) return value < i.value;
return id < i.id;
}
bool operator>(const Item i) const {
if (value != i.value) return value > i.value;
return id > i.id;
}
};
ostream& operator<<(ostream& out, Item const& item) {
return out << "(" << item.id << ':' << item.value << ")";
}
class MBest {
long long best_sum;
multiset<int, greater<int> > le;
multiset<int> gt;
void transfer_gt_to_le() {
assert(!gt.empty());
auto it = gt.begin();
le.insert(*it);
best_sum += *it;
gt.erase(it);
}
void transfer_le_to_gt() {
assert(!le.empty());
auto it = le.begin();
gt.insert(*it);
best_sum -= *it;
le.erase(it);
}
public:
MBest() : best_sum(0) {}
int getM() const { return le.size(); }
long long query_best_sum() const { return best_sum; }
void add(int x) {
le.insert(x);
best_sum += x;
transfer_le_to_gt();
}
void remove(int x) {
auto it_gt = gt.find(x);
if (it_gt != gt.end()) {
gt.erase(it_gt);
return;
}
auto it_le = le.find(x);
assert(it_le != le.end());
le.erase(it_le);
best_sum -= x;
transfer_gt_to_le();
}
void increaseM() { transfer_gt_to_le(); }
};
vector<long long> get_psum(const vector<Item>& V) {
vector<long long> res(int((V).size()));
for (int i = (1), _b = (int((V).size()) - 1); i <= _b; ++i)
res[i] = res[i - 1] + V[i].value;
return res;
}
long long solve(int N, int M, int K, const vector<int>& T, const vector<int>& A,
const vector<int>& B) {
MBest mb;
vector<Item> alice, bob, both;
for (int i = 0, _n = (N); i < _n; ++i) {
Item item = {i + 1, T[i]};
if (A[i] && B[i])
both.push_back(item);
else if (A[i])
alice.push_back(item);
else if (B[i])
bob.push_back(item);
else
mb.add(item.value);
}
int minj = min(int((alice).size()), int((bob).size()));
if (minj + int((both).size()) < K) {
cout << "-1" << '\n';
return -1;
}
sort(alice.begin(), alice.end());
sort(bob.begin(), bob.end());
sort(both.begin(), both.end());
minj = min(minj, K);
for (int j = (minj), _b = (int((alice).size()) - 1); j <= _b; ++j)
mb.add(alice[j].value);
alice.erase(alice.begin() + minj, alice.end());
alice.insert(alice.begin(), {0, 0});
for (int j = (minj), _b = (int((bob).size()) - 1); j <= _b; ++j)
mb.add(bob[j].value);
bob.erase(bob.begin() + minj, bob.end());
bob.insert(bob.begin(), {0, 0});
both.insert(both.begin(), {0, 0});
for (int k = (K - minj), _b = (int((both).size()) - 1); k <= _b; ++k)
mb.add(both[k].value);
vector<long long> psum_alice = get_psum(alice);
vector<long long> psum_bob = get_psum(bob);
vector<long long> psum_both = get_psum(both);
long long res = INF;
int bestj = -1;
for (int j = minj; j >= 0; --j) {
int k = K - j;
if (k >= int((both).size())) break;
mb.remove(both[k].value);
int nbooks = K + j;
if (nbooks <= M) {
int nmissing = M - nbooks;
while (mb.getM() < nmissing) mb.increaseM();
long long best_other = mb.query_best_sum();
long long cur = psum_alice[j] + psum_bob[j] + psum_both[k] + best_other;
if (res > cur) {
bestj = j;
res = cur;
}
}
if (j > 0) {
mb.add(alice[j].value);
mb.add(bob[j].value);
}
}
if (bestj < 0) {
cout << "-1\n";
return -1;
}
set<Item> avail;
for (int i = 0, _n = (N); i < _n; ++i) avail.insert({i + 1, T[i]});
vector<int> best_books;
for (int j = (1), _b = (bestj); j <= _b; ++j) {
Item item = alice[j];
avail.erase(item);
best_books.push_back(item.id);
}
for (int j = (1), _b = (bestj); j <= _b; ++j) {
Item item = bob[j];
avail.erase(item);
best_books.push_back(item.id);
}
for (int k = (1), _b = (K - bestj); k <= _b; ++k) {
Item item = both[k];
avail.erase(item);
best_books.push_back(item.id);
}
assert(int((best_books).size()) <= M);
while (int((best_books).size()) < M) {
auto it = avail.begin();
best_books.push_back(it->id);
avail.erase(it);
}
cout << res << '\n';
long long sum = 0;
for (int id : best_books) {
cout << id << ' ';
sum += T[id - 1];
}
cout << '\n';
return res;
}
int main(int argc, char* argv[]) {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int N, M, K;
cin >> N >> M >> K;
vector<int> T(N), A(N), B(N);
for (int i = 0, _n = (N); i < _n; ++i) cin >> T[i] >> A[i] >> B[i];
solve(N, M, K, T, A, B);
return 0;
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
class Solve {
public:
void updateSt(set<pair<int, int>> &st, set<pair<int, int>> &fr, int &sum,
int need) {
need = max(need, 0);
while (true) {
bool useful = false;
while (st.size() > need) {
sum -= st.rbegin()->first;
fr.insert(*st.rbegin());
st.erase(prev(st.end()));
useful = true;
}
while (st.size() < need && fr.size() > 0) {
sum += fr.begin()->first;
st.insert(*fr.begin());
fr.erase(fr.begin());
useful = true;
}
while (!st.empty() && !fr.empty() &&
fr.begin()->first < st.rbegin()->first) {
sum -= st.rbegin()->first;
sum += fr.begin()->first;
fr.insert(*st.rbegin());
st.erase(prev(st.end()));
st.insert(*fr.begin());
fr.erase(fr.begin());
useful = true;
}
if (!useful) break;
}
}
void main() {
const int INF = 10e9;
int n, m, k;
cin >> n >> m >> k;
vector<pair<int, int>> times[4];
vector<int> sums[4];
for (int i = 0; i < n; ++i) {
int t, a, b;
cin >> t >> a >> b;
times[a * 2 + b].push_back({t, i});
}
for (int i = 0; i < 4; ++i) {
sort(times[i].begin(), times[i].end());
sums[i].push_back(0);
for (auto it : times[i]) {
sums[i].push_back(sums[i].back() + it.first);
}
}
int ans = INF;
int pos = INF;
set<pair<int, int>> st;
set<pair<int, int>> fr;
int sum = 0;
vector<int> res;
for (int iter = 0; iter < 2; ++iter) {
st.clear();
fr.clear();
sum = 0;
int start = 0;
while (k - start >= sums[1].size() || k - start >= sums[2].size() ||
m - start - (k - start) * 2 < 0) {
++start;
}
if (start >= sums[3].size()) {
cout << -1 << endl;
return;
}
int need = m - start - (k - start) * 2;
for (int i = 0; i < 3; ++i) {
for (int p = times[i].size() - 1; p >= (i == 0 ? 0 : k - start); --p) {
fr.insert(times[i][p]);
}
}
updateSt(st, fr, sum, need);
for (int cnt = start; cnt < (iter == 0 ? sums[3].size() : pos); ++cnt) {
if (k - cnt >= 0) {
if (cnt + (k - cnt) * 2 + st.size() == m) {
if (ans >
sums[3][cnt] + sums[1][k - cnt] + sums[2][k - cnt] + sum) {
ans = sums[3][cnt] + sums[1][k - cnt] + sums[2][k - cnt] + sum;
pos = cnt + 1;
}
}
} else {
if (cnt + st.size() == m) {
if (ans > sums[3][cnt] + sum) {
ans = sums[3][cnt] + sum;
pos = cnt + 1;
}
}
}
if (iter == 1 && cnt + 1 == pos) break;
need -= 1;
if (k - cnt > 0) {
need += 2;
fr.insert(times[1][k - cnt - 1]);
fr.insert(times[2][k - cnt - 1]);
}
updateSt(st, fr, sum, need);
}
if (iter == 1) {
for (int i = 0; i + 1 < pos; ++i) res.push_back(times[3][i].second);
for (int i = 0; i <= k - pos; ++i) {
res.push_back(times[1][i].second);
res.push_back(times[2][i].second);
}
for (auto [value, position] : st) res.push_back(position);
}
}
cout << ans << endl;
for (auto it : res) cout << it + 1 << " ";
cout << endl;
return;
}
};
int main(int argc, const char *argv[]) {
Solve s;
s.main();
return 0;
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python2
|
input = raw_input
range = xrange
import sys
inp = [int(x) for x in sys.stdin.read().split()]; ii = 0
seg = [0]*200000
def offset(x):
return x + 100000
def encode(x, y):
return (x<<20) + y
def upd(node, L, R, pos, val):
if L+1 == R:
seg[node] += val
seg[offset(node)] = seg[node]*L
return
M = (L+R)>>1
if pos < M:
upd(node<<1, L, M, pos, val)
else:
upd(node<<1 | 1, M, R, pos, val)
seg[node] = seg[node<<1] + seg[node<<1 | 1]
seg[offset(node)] = seg[offset(node<<1)] + seg[offset(node<<1 | 1)]
def query(node, L, R, k):
if k == 0:
return [0, 0]
if seg[node] < k:
return seg[offset(node)], seg[node]
if L+1 == R:
return [L*k, k]
M = (L+R)>>1
leftval, leftct = query(node<<1, L, M, k)
rightval, rightct = query(node<<1 | 1, M, R, k-leftct)
return leftval+rightval, leftct+rightct
n, m, k = inp[ii:ii+3]; ii += 3
A, B, AB, notAB = [], [], [], []
for i in range(n):
t, a, b = inp[ii:ii+3]; ii += 3
if a == 0 and b == 0:
notAB.append(encode(t, i+1))
if a == 1 and b == 0:
A.append(encode(t, i+1))
if a == 0 and b == 1:
B.append(encode(t, i+1))
if a == 1 and b == 1:
AB.append(encode(t, i+1))
upd(1, 0, 10001, t, 1)
A.sort(); B.sort(); AB.sort()
p1 = min(k, len(AB))
p2 = k - p1
if 2*k - p1 > m or p2 > min(len(A), len(B)):
print(-1)
exit(0)
sum, ans, ch = 0, 2**31, p1
for i in range(p1):
sum += AB[i]>>20
upd(1, 0, 10001, AB[i]>>20, -1)
for i in range(p2):
sum += (A[i]>>20) + (B[i]>>20)
upd(1, 0, 10001, A[i]>>20, -1)
upd(1, 0, 10001, B[i]>>20, -1)
ans, _ = query(1, 0, 10001, m-2*k+p1)
ans += sum
while p1 > 0:
if p2 == min(len(A), len(B)):
break
upd(1, 0, 10001, A[p2]>>20, -1); sum += A[p2]>>20
upd(1, 0, 10001, B[p2]>>20, -1); sum += B[p2]>>20
upd(1, 0, 10001, AB[p1-1]>>20, 1); sum -= AB[p1-1]>>20
p2 += 1
p1 -= 1
if m - 2*k + p1 < 0:
break
Q, _ = query(1, 0, 10001, m-2*k+p1)
if ans > sum + Q:
ans = sum + Q
ch = p1
print ans
ind = [AB[i]&((1<<20)-1) for i in range(ch)] + [A[i]&((1<<20)-1) for i in range(k-ch)] + [B[i]&((1<<20)-1) for i in range(k-ch)]
st = sorted(notAB + AB[ch:] + A[k-ch:] + B[k-ch:])
ind += [st[i]&((1<<20)-1) for i in range(m-2*k+ch)]
print ' '.join(str(x) for x in ind)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
## necessary imports
import sys
input = sys.stdin.readline
#from math import ceil, floor, factorial;
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp
## gcd function
def gcd(a,b):
if b == 0:
return a
return gcd(b, a % b);
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if(k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return int(res)
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0):
hi = len(a)
while lo < hi:
mid = (lo+hi)//2
if a[mid] < x:
lo = mid+1
else:
hi = mid
return lo
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0 and n > 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0 and n > 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve()
def factoriazation(x):
ret = {};
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1;
x = x//spf[x]
return ret
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()))
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
n, k = int_array(); a = [];
sum_a = sum_b = 0;
for __ in range(n):
t, aa, bb = int_array();
sum_a += aa; sum_b += bb;
a.append((t, aa, bb));
if sum_a < k or sum_b < k:
print(-1);
else:
only_a = []; only_b = []; both = [];
for i in a:
if i[1] == 1 and i[2] == 0:
only_a.append(i[0]);
elif i[1] == 0 and i[2] == 1:
only_b.append(i[0]);
elif i[1] == 1 and i[2] == 1:
both.append(i[0]);
only_a.sort(); only_b.sort();
for i in range(min(len(only_a), len(only_b))):
both.append(only_a[i] + only_b[i]);
both.sort();
print(sum(both[:k]));
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
######################################################
############Created by Devesh Kumar###################
#############devesh1102@gmail.com####################
##########For CodeForces(Devesh1102)#################
#####################2020#############################
######################################################
import sys
input = sys.stdin.readline
# import sys
import heapq
import copy
import math
import decimal
# import sys.stdout.flush as flush
# from decimal import *
#heapq.heapify(li)
#
#heapq.heappush(li,4)
#
#heapq.heappop(li)
#
# & Bitwise AND Operator 10 & 7 = 2
# | Bitwise OR Operator 10 | 7 = 15
# ^ Bitwise XOR Operator 10 ^ 7 = 13
# << Bitwise Left Shift operator 10<<2 = 40
# >> Bitwise Right Shift Operator
# '''############ ---- Input Functions ---- #######Start#####'''
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def insr2():
s = input()
return((s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
############ ---- Input Functions ---- #######End
# #####
def pr_list(a):
print(*a, sep=" ")
def main():
# tests = inp()
tests = 1
# mod = 1000000007
limit = 10**18
ans = 0
for test in range(tests):
[n,k] = inlt()
b = []
a = []
both = []
for i in range(n):
[t,alice,bob] = inlt()
if alice == 1 and bob == 1:
both.append(t)
elif alice == 1:
a.append(t)
elif bob == 1:
b.append(t)
both.sort()
a.sort()
b.sort()
if len(both) + len(a) <k or len(both) + len(b) <k:
print(-1)
break
# print(both,a,b)
ans = 0
a_i = 0
b_i = 0
both_i = 0
for _ in range(k):
if both_i != len(both) and a_i !=len(a) and b_i!= len(b):
if both[both_i] < a[a_i] + b[b_i]:
ans = ans + both[both_i]
both_i +=1
else:
ans = ans + a[a_i] + b[b_i]
a_i = a_i +1
b_i = b_i +1
elif both_i == len(both):
ans = ans + a[a_i] + b[b_i]
a_i = a_i +1
b_i = b_i +1
else:
ans = ans + both[both_i]
both_i +=1
print(ans)
if __name__== "__main__":
main()
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
n, k = map(int, input().split())
A, B, both = [0], [0], [0]
for i in range(n):
t, a, b = map(int, input().split())
if a == 1 and b == 1:
both.append(t)
if a == 1 and b == 0:
A.append(t)
if a == 0 and b == 1:
B.append(t)
A.sort()
B.sort()
both.sort()
if len(A)-1 + len(both)-1 < k or len(B)-1 + len(both)-1 < k:
print(-1)
exit(0)
for i in range(1, len(A)):
A[i] += A[i-1]
for i in range(1, len(B)):
B[i] += B[i-1]
cur = 0
ans = 10**12 + 10
for ch in range(len(both)):
cur += both[ch]
if ch > k:
break
# Need k - ch each from A and B
if k - ch < len(A) and k - ch < len(B):
ans = min(ans, A[k-ch] + B[k-ch] + cur)
print(ans)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
line = input().split()
n,k = int(line[0]), int(line[1])
a,b,c = [],[],[]
for i in range(n):
line = input().split()
t,ai,bi = int(line[0]), int(line[1]), int(line[2])
if (ai == 1 and bi == 1):
c.append(t)
elif (ai == 1):
a.append(t)
elif (bi == 1):
b.append(t)
a.sort()
b.sort()
c.sort()
i,j,cnt = 0,0,0
ans = 0
while (i < min(len(a),len(b)) and j < len(c) and cnt < k):
if (a[i]+b[i] < c[j]):
ans += a[i]+b[i]
i += 1
else:
ans += c[j]
j += 1
cnt += 1
while (i < min(len(a),len(b)) and cnt < k):
ans += a[i]+b[i]
cnt += 1
i += 1
while (j < len(c) and cnt < k):
ans += c[j]
cnt += 1
j += 1
if (cnt < k):
print(-1)
else:
print(ans)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.InputMismatchException;
public class B {
public static void main(String[] args) throws Exception {
// TODO Auto-generated method stub
InputReader s = new InputReader(System.in);
PrintWriter p = new PrintWriter(System.out);
int n = s.nextInt();
int k = s.nextInt();
ArrayList<Integer> alice = new ArrayList<>();
ArrayList<Integer> bob = new ArrayList<>();
ArrayList<Integer> both = new ArrayList<>();
int atotal = 0, btotal = 0;
for (int i = 0; i < n; i++) {
int val = s.nextInt();
int a = s.nextInt();
int b = s.nextInt();
if (a * b == 1) {
both.add(val);
atotal++;
btotal++;
} else if (a == 1) {
atotal++;
alice.add(val);
} else if (b == 1) {
btotal++;
bob.add(val);
}
}
if (atotal < k || btotal < k) {
p.println(-1);
p.flush();
return;
}
Collections.sort(alice, Collections.reverseOrder());
Collections.sort(bob, Collections.reverseOrder());
Collections.sort(both, Collections.reverseOrder());
long ans = 0;
for (int i = 0; i < k; i++) {
long val1 = Integer.MAX_VALUE;
long val2 = Integer.MAX_VALUE;
if (alice.size() > 0 && bob.size() > 0) {
val1 = alice.get(alice.size() - 1) + bob.get(bob.size() - 1);
}
if (both.size()>0) {
val2 = both.get(both.size() - 1);
}
if (val1 > val2) {
ans += val2;
both.remove(both.size() - 1);
} else {
ans += val1;
alice.remove(alice.size() - 1);
bob.remove(bob.size() - 1);
}
}
p.println(ans);
p.flush();
p.close();
}
public static boolean prime(long n) {
if (n == 1) {
return false;
}
if (n == 2) {
return true;
}
for (long i = 2; i <= (long) Math.sqrt(n); i++) {
if (n % i == 0)
return false;
}
return true;
}
public static ArrayList Divisors(long n) {
ArrayList<Long> div = new ArrayList<>();
for (long i = 1; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
div.add(i);
if (n / i != i)
div.add(n / i);
}
}
return div;
}
public static int BinarySearch(long[] a, long k) {
int n = a.length;
int i = 0, j = n - 1;
int mid = 0;
if (k < a[0])
return 0;
else if (k >= a[n - 1])
return n;
else {
while (j - i > 1) {
mid = (i + j) / 2;
if (k >= a[mid])
i = mid;
else
j = mid;
}
}
return i + 1;
}
public static long GCD(long a, long b) {
if (b == 0)
return a;
else
return GCD(b, a % b);
}
public static long LCM(long a, long b) {
return (a * b) / GCD(a, b);
}
static class pair implements Comparable<pair> {
Long x, y;
pair(long x, long y) {
this.x = x;
this.y = y;
}
public int compareTo(pair o) {
int result = x.compareTo(o.x);
if (result == 0)
result = y.compareTo(o.y);
return result;
}
public String toString() {
return x + " " + y;
}
public boolean equals(Object o) {
if (o instanceof pair) {
pair p = (pair) o;
return p.x - x == 0 && p.y - y == 0;
}
return false;
}
public int hashCode() {
return new Long(x).hashCode() * 31 + new Long(y).hashCode();
}
}
static class InputReader {
private final InputStream stream;
private final byte[] buf = new byte[8192];
private int curChar, snumChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int snext() {
if (snumChars == -1)
throw new InputMismatchException();
if (curChar >= snumChars) {
curChar = 0;
try {
snumChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (snumChars <= 0)
return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public int[] nextIntArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
return a;
}
public String readString() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isSpaceChar(c));
return res.toString();
}
public String nextLine() {
int c = snext();
while (isSpaceChar(c))
c = snext();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isEndOfLine(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
static class CodeX {
public static void sort(long arr[]) {
merge_sort(arr, 0, arr.length - 1);
}
private static void merge_sort(long A[], long start, long end) {
if (start < end) {
long mid = (start + end) / 2;
merge_sort(A, start, mid);
merge_sort(A, mid + 1, end);
merge(A, start, mid, end);
}
}
private static void merge(long A[], long start, long mid, long end) {
long p = start, q = mid + 1;
long Arr[] = new long[(int) (end - start + 1)];
long k = 0;
for (int i = (int) start; i <= end; i++) {
if (p > mid)
Arr[(int) k++] = A[(int) q++];
else if (q > end)
Arr[(int) k++] = A[(int) p++];
else if (A[(int) p] < A[(int) q])
Arr[(int) k++] = A[(int) p++];
else
Arr[(int) k++] = A[(int) q++];
}
for (int i = 0; i < k; i++) {
A[(int) start++] = Arr[i];
}
}
}
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
n,k=map(int,input().split(" "))
db=[]
a=[]
b=[]
for i in range(n):
t,A,B= map(int, input().split(" "))
if A==B==1:
db.append(t)
elif A==1:
a.append(t)
elif B==1:
b.append(t)
a.sort()
b.sort()
for i in range(min(len(a),len(b))):
db.append((a[i]+b[i]))
if len(db)<k:
print(-1)
else:
db.sort()
print(sum(db[:k]))
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python2
|
n,k = map(int,raw_input().split())
both = [];alice = [];bob = []
for _ in range(n):
t,a,b = map(int,raw_input().split())
if a&b:
both.append(t)
elif a:
alice.append(t)
elif b:
bob.append(t)
alice.sort()
bob.sort()
for i in range(min(len(alice),len(bob))):
both.append(alice[i]+bob[i])
print(-1if len(both)<k else sum(sorted(both)[:k]))
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
//package com.company;
//import com.sun.tools.corba.se.idl.StructEntry;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
String s1 = read.readLine();
String s[] = s1.split(" ");
int count=0;
int t = Integer.parseInt(s[0]);
int k = Integer.parseInt(s[1]);
PriorityQueue<Integer> a = new PriorityQueue<Integer>();
PriorityQueue<Integer> b = new PriorityQueue<Integer>();
PriorityQueue<Integer> ab = new PriorityQueue<Integer>();
for (int i = 0; i < t; i++) {
String s2 = read.readLine();
String s3[] = s2.split(" ");
int time=Integer.parseInt(s3[0]);
int A=Integer.parseInt(s3[1]);
int B=Integer.parseInt(s3[2]);
if(A==1&&B==1){
ab.add(time);
count++;
}
else if(A==1)
{a.add(time);
count++;
}
else if(B==1){
b.add(time);
count++;
}
}
if((a.size()+ab.size())<k||(b.size()+ab.size())<k)
System.out.println("-1");
else{
int countA=0;
int countB=0;
int tt=0;
while(countA<k&&countB<k){
if(countA<k&&countB<k){
if(ab.isEmpty()){
tt+=a.poll();
tt+=b.poll();
countA++;
countB++;
}
else{
if(a.isEmpty()){
tt+=ab.poll();
countA++;
countB++;
}
else if(b.isEmpty()){
tt+=ab.poll();
countA++;
countB++;
}
else{
if(ab.peek()>a.peek()+b.peek()){
tt+=a.poll();
tt+=b.poll();
countA++;
countB++;
}
else{
tt+=ab.poll();
countA++;
countB++;
}
}
}
}else if(countB<k){
if(ab.isEmpty()){
tt+=b.poll();
countB++;
}
else{
if(b.isEmpty()){
tt+=ab.poll();
countA++;
countB++;
}
else{
if(b.peek()<ab.peek()){
tt+=b.poll();
countB++;
}else{
tt+=ab.poll();
countA++;
countB++;
}
}
}
}
else if(countA<k){
if(ab.isEmpty()){
tt+=a.poll();
countA++;
}
else{
if(a.isEmpty()){
tt+=ab.poll();
countA++;
countB++;
}
else{
if(a.peek()<ab.peek()){
tt+=a.poll();
countA++;
}else{
tt+=ab.poll();
countA++;
countB++;
}
}
}
}
}
System.out.println(tt);
}
}
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
import os, sys
from io import IOBase, BytesIO
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = 'x' in file.mode or 'w' in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO,self).read()
def readline(self):
while self.newlines == 0:
s = self._fill(); self.newlines = s.count(b'\n') + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
self.write = lambda s:self.buffer.write(s.encode('ascii'))
self.read = lambda:self.buffer.read().decode('ascii')
self.readline = lambda:self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
# Cout implemented in Python
import sys
class ostream:
def __lshift__(self,a):
sys.stdout.write(str(a))
return self
cout = ostream()
endl = '\n'
def get_input(a=str):
return a(input())
def get_int_input():
return get_input(int)
def get_input_arr(a):
return list(map(a, input().split()))
def get_int_input_arr():
return get_input_arr(int)
def solve():
n, k = get_int_input_arr()
books_both = []
books_a = []
books_b = []
for _ in range(n):
t_i, a_i, b_i = get_int_input_arr()
if a_i == 1 and b_i == 1:
books_both.append(t_i)
elif a_i == 1 and b_i == 0:
books_a.append(t_i)
elif a_i == 0 and b_i == 1:
books_b.append(t_i)
books_both.sort()
books_a.sort()
books_b.sort()
prefx_both = [0] * (len(books_both) + 1)
for i in range(1, len(books_both) + 1):
prefx_both[i] = prefx_both[i - 1] + books_both[i - 1]
prefx_a = [0] * (len(books_a) + 1)
for i in range(1, len(books_a) + 1):
prefx_a[i] = prefx_a[i - 1] + books_a[i - 1]
prefx_b = [0] * (len(books_b) + 1)
for i in range(1, len(books_b) + 1):
prefx_b[i] = prefx_b[i - 1] + books_b[i - 1]
# print(books_both)
def can_do(time):
for i in range(min(k + 1, len(books_both) + 1)):
# print(i)
both_books_time = prefx_both[i]
left_nums = k - i
if left_nums < 0 or left_nums >= len(prefx_a) or left_nums >= len(prefx_b):
continue
books_a_time = prefx_a[left_nums]
books_b_time = prefx_b[left_nums]
if time >= both_books_time + books_a_time + books_b_time:
return True
return False
lo = 0
hi = 10 ** 10
res = float("inf")
while lo <= hi:
mid = lo + (hi - lo) // 2
if can_do(mid):
res = mid
hi = mid - 1
else:
lo = mid + 1
if res == float("inf"):
cout<<-1<<endl
else:
cout<<res<<endl
def main():
solve()
if __name__ == "__main__":
main()
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
n,k=map(int,input().split())
a =[]
b =[]
c = []
for i in range(n):
m = [*map(int,input().split())]
if(m[1]==1 and m[2]==1):
c.append(m[0])
elif(m[1]==1):
a.append(m)
elif(m[2]==1):
b.append(m)
a.sort(key=lambda x:(x[0]))
b.sort(key=lambda x:(x[0]))
for i in range(min(len(a),len(b))):
c.append(a[i][0]+b[i][0])
c.sort()
if(len(c)<k):
print(-1)
else:
print(sum(c[i] for i in range(k)))
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
import sys
input=sys.stdin.readline
def main():
n,k=map(int,input().split())
A,B,C=[],[],[]
for i in range(n):
n,a,b=map(int,input().split())
if(a==1 and b==1):
A.append(n)
elif(a==1 and b==0):
B.append(n)
elif(a==0 and b==1):
C.append(n)
B.sort()
C.sort()
v=min(len(B),len(C))
for i in range(v):
A.append(B[i]+C[i])
if(len(A)<k):
print(-1)
else:
A.sort()
ans=0
for i in range(k):
ans=ans+A[i]
print(ans)
main()
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
int n, m, k;
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cin >> n >> m >> k;
vector<pair<int, int> > ab;
vector<pair<int, int> > a;
vector<pair<int, int> > b;
vector<pair<int, int> > non;
for (int i = 0; i < n; i++) {
int t, a1, b1;
cin >> t >> a1 >> b1;
if (a1 && b1) {
ab.push_back({t, i});
} else if (a1) {
a.push_back({t, i});
} else if (b1) {
b.push_back({t, i});
} else {
non.push_back({t, i});
}
}
sort(ab.begin(), ab.end());
sort(a.begin(), a.end());
sort(b.begin(), b.end());
sort(non.begin(), non.end());
int l = -1;
for (int i = 0; i <= min(k, (int)ab.size()); i++) {
if (k - i <= (int)a.size() && k - i <= (int)b.size() &&
((k - i) * 2 + i <= m)) {
l = i;
break;
}
}
if (l == -1) {
cout << -1;
return 0;
}
set<pair<int, int> > curr;
set<pair<int, int> > all;
int r1 = l, r2 = k - l;
long long ans;
long long curr_sum = 0;
int ans_r1 = r1, ans_r2 = r2;
int curr_el = r2 * 2 + r1;
for (int i = 0; i < r2; i++) {
curr_sum += a[i].first;
curr_sum += b[i].first;
}
for (int i = 0; i < r1; i++) {
curr_sum += ab[i].first;
}
for (int i = r2; i < a.size(); i++) {
all.insert(a[i]);
}
for (int i = r2; i < b.size(); i++) {
all.insert(b[i]);
}
for (int i = 0; i < non.size(); i++) {
all.insert(non[i]);
}
for (int i = r1; i < ab.size(); i++) {
all.insert(ab[i]);
}
int add = 0;
while (curr.size() < (m - curr_el)) {
curr_sum += (*all.begin()).first;
curr.insert(*all.begin());
all.erase(all.begin());
}
ans = curr_sum;
for (int i = l; i < min(k, (int)ab.size()); i++) {
curr_sum -= a[r2 - 1].first;
curr_sum -= b[r2 - 1].first;
curr_sum += ab[r1].first;
all.insert(a[r2 - 1]);
all.insert(b[r2 - 1]);
r2--;
if (all.find(ab[r1]) != all.end()) {
all.erase(ab[r1]);
}
if (curr.find(ab[r1]) != curr.end()) {
curr_sum -= ab[r1].first;
curr.erase(ab[r1]);
}
r1++;
curr_el = r2 * 2 + r1;
while (curr.size() < (m - curr_el)) {
curr_sum += (*all.begin()).first;
curr.insert(*all.begin());
all.erase(all.begin());
}
if (!curr.empty() && !all.empty()) {
auto curr_last = curr.end();
curr_last--;
while (!all.empty() && *curr_last > *all.begin()) {
curr_sum -= (*curr_last).first;
curr_sum += (*all.begin()).first;
all.insert(*curr_last);
curr.erase(curr_last);
curr.insert(*all.begin());
all.erase(all.begin());
curr_last = curr.end();
curr_last--;
}
}
if (curr_sum < ans) {
ans = curr_sum;
ans_r1 = r1;
ans_r2 = r2;
}
}
r1 = ans_r1;
r2 = ans_r2;
cout << ans << "\n";
for (int i = 0; i < r1; i++) {
cout << ab[i].second + 1 << " ";
}
for (int i = 0; i < r2; i++) {
cout << a[i].second + 1 << " ";
cout << b[i].second + 1 << " ";
}
set<pair<int, int> > free;
for (int i = r2; i < a.size(); i++) {
free.insert(a[i]);
}
for (int i = r2; i < b.size(); i++) {
free.insert(b[i]);
}
for (int i = 0; i < non.size(); i++) {
free.insert(non[i]);
}
for (int i = r1; i < ab.size(); i++) {
free.insert(ab[i]);
}
curr_el = r2 * 2 + r1;
for (int i = 0; i < (m - curr_el); i++) {
cout << (*free.begin()).second + 1 << " ";
free.erase(free.begin());
}
return 0;
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
from collections import defaultdict as dd
import sys
input=sys.stdin.readline
#n=int(input())
n,kk=map(int,input().split())
l=[]
ans=0
for i in range(n):
time,a,b=map(int,input().split())
l.append((time,a,b))
l.sort()
la=[]
lb=[]
do=[]
for i in l:
time,a,b=i
if(a and b):
do.append(time)
elif(a):
la.append(time)
elif(b):
lb.append(time)
i=0
j=0
k=0
ca=kk
cb=kk
while ca and cb:
if(i<len(la)):
if(j<len(lb)):
if(k<len(do)):
if(la[i]+lb[j]<=do[k]):
ans+=la[i]
ans+=lb[j]
i+=1
j+=1
ca-=1
cb-=1
else:
ans+=do[k]
k+=1
ca-=1
cb-=1
else:
ans+=la[i]
ans+=lb[j]
i+=1
j+=1
ca-=1
cb-=1
else:
if(k<len(do)):
ans+=do[k]
k+=1
ca-=1
cb-=1
else:
break
else:
if(k<len(do)):
ans+=do[k]
k+=1
ca-=1
cb-=1
else:
break
if(ca):
while ca and i<len(la) and k<len(do):
if(la[i]<=do[k]):
ans+=la[i]
i+=1
ca-=1
else:
ans+=do[k]
k+=1
ca-=1
while ca and i<len(la):
ans+=la[i]
i+=1
ca-=1
while ca and k<len(do):
ans+=do[k]
k+=1
ca-=1
elif(cb):
i=j
la=lb
ca=cb
while ca and i<len(la) and k<len(do):
if(la[i]<=do[k]):
ans+=la[i]
i+=1
ca-=1
else:
ans+=do[k]
k+=1
ca-=1
while ca and i<len(la):
ans+=la[i]
i+=1
ca-=1
while ca and k<len(do):
ans+=do[k]
k+=1
ca-=1
cb=ca
if(ca or cb):
print(-1)
else:
print(ans)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int mpow(int base, int exp);
void ipgraph(int n, int m);
void dfs(int u, int par);
const int mod = 1e9 + 7;
struct cmp {
bool operator()(const int& lhs, const int& rhs) const { return lhs < rhs; }
};
int mpow(int base, int exp) {
base %= mod;
int result = 1;
while (exp > 0) {
if (exp & 1) result = ((long long)result * base) % mod;
base = ((long long)base * base) % mod;
exp >>= 1;
}
return result;
}
bool isPrime(int n) {
if (n <= 1) return false;
if (n <= 3) return true;
if (n % 2 == 0 || n % 3 == 0) return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0) return false;
return true;
}
void solve() {
long long n, k, time;
bool l1, l2;
cin >> n >> k;
vector<int> a1, a2, a3;
for (int i = 0; i < n; i++) {
cin >> time;
cin >> l1 >> l2;
if (l1 & l2) {
a1.push_back(time);
} else if (l1) {
a2.push_back(time);
} else if (l2) {
a3.push_back(time);
}
}
if (a1.size() + a2.size() < k || a1.size() + a3.size() < k) {
cout << -1;
return;
}
sort(a1.begin(), a1.end());
sort(a2.begin(), a2.end());
sort(a3.begin(), a3.end());
int p1 = 0, p2 = 0, p3 = 0;
int cnt = 0, res = 0;
int s1 = a1.size(), s2 = a2.size(), s3 = a3.size();
while (cnt < k && p1 < s1 && p2 < s2 && p3 < s3) {
if (a1[p1] < a2[p2] + a3[p3]) {
res += a1[p1];
p1++;
} else {
res += a2[p2] + a3[p3];
p2++;
p3++;
}
cnt++;
}
while (cnt < k && p1 < s1) {
res += a1[p1];
p1++;
cnt++;
}
while (cnt < k && p2 < s2 && p3 < s3) {
res += a2[p2] + a3[p3];
p2++;
p3++;
cnt++;
}
cout << res;
}
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int t = 1;
solve();
cout << "\n";
return 0;
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
import sys
from collections import deque, defaultdict
input = sys.stdin.buffer.readline
n, k = map(int, input().split())
al, bl, cc = [], [], []
for _ in range(n):
t, a, b = map(int, input().split())
if a and not b: al.append(t)
elif not a and b: bl.append(t)
elif a and b: cc.append(t)
al.sort(); bl.sort()
for i in range(min(len(al), len(bl))):
cc.append(al[i]+bl[i])
cc.sort()
if len(cc) < k: print(-1)
else: print(sum(cc[0:k]))
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
n,k=map(int, input().split())
tab=[tuple(map(int,input().split())) for _ in range(n)]
both=[tabi[0] for tabi in tab if tabi[1:]==(1,1)]
both.sort()
alice=[tabi[0] for tabi in tab if tabi[1:]==(1,0)]# tabi[1]==1 and tabi[2]==0]
bob=[tabi[0] for tabi in tab if tabi[1:]==(0,1)]
alice.sort()
bob.sort()
ab=[alice[i]+bob[i] for i in range(min(len(alice),len(bob)))]
both+=ab
both.sort()
if len(both)<k:
print(-1)
else:print(sum(both[:k]))
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
n,k = [int(x) for x in input().strip().split()]
vec = []
for _ in range(n):
vec.append([int(x) for x in input().split()])
vec.sort()
f1 = f2 = f3 = 0
minsum = 0
count = 0
min_12 = min_3 = 0
flag12 = flag3 = True
while ((f3 <= n-1 or (f1 <= n-1 and f2 <= n-1)) and count < k):
while f3 <= n-1:
if vec[f3][1] == 1 and vec[f3][2] == 1:
break
f3 += 1
if f3 <= n-1:
min_3 = vec[f3][0]
else:
flag3 = False
while f2 <= n-1:
if vec[f2][1] == 1 and vec[f2][2] == 0:
break
f2 += 1
while f1 <= n-1:
if vec[f1][1] == 0 and vec[f1][2] == 1:
break
f1 += 1
if f1 <= n-1 and f2 <= n-1:
min_12 = vec[f1][0] + vec[f2][0]
else:
flag12 = False
if flag12 and flag3:
count +=1
if min_3 <= min_12:
f3 += 1
minsum += min_3
else:
f1 += 1
f2 += 1
minsum += min_12
elif flag12:
count += 1
f1 += 1
f2 += 1
minsum += min_12
elif flag3:
count += 1
f3 += 1
minsum += min_3
if count == k:
print(minsum)
else:
print(-1)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
n,k=map(int,input().split())
l1=[]
l2=[]
l3=[]
ac,bc=0,0
for i in range(n):
t,a,b=map(int,input().split())
if(a==1 and b==1):
l1.append(t)
ac+=1
bc+=1
elif(a==1 and b==0):
l2.append(t)
ac+=1
elif(a==0 and b==1):
l3.append(t)
bc+=1
if(ac<k or bc<k):
print(-1)
else:
l2.sort()
l3.sort()
ans=0
for i in range(min(len(l2),len(l3))):
l1.append(l2[i]+l3[i])
l1.sort()
for i in range(k):
ans+=l1[i]
print(ans)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
vector<pair<int, pair<int, int>>> v;
for (int i = 0; i < n; i++) {
int t, a, b;
cin >> t >> a >> b;
v.push_back({t, {a, b}});
}
vector<int> f;
vector<pair<int, pair<int, int>>> l;
vector<pair<int, pair<int, int>>> r;
for (auto xx : v) {
if ((xx.second.second == 1) && (xx.second.first == 1)) {
f.push_back(xx.first);
}
if ((xx.second.second == 0) && (xx.second.first == 1)) {
l.push_back(xx);
}
if ((xx.second.second == 1) && (xx.second.first == 0)) {
r.push_back(xx);
}
}
sort(l.begin(), l.end());
sort(r.begin(), r.end());
int id1 = 0, id2 = 0;
while ((id1 < l.size()) && (id2 < r.size())) {
f.push_back(l[id1].first + r[id2].first);
id1++, id2++;
}
sort(f.begin(), f.end());
long long sum = 0;
if (f.size() < k) {
cout << "-1"
<< "\n";
return 0;
}
for (int i = 0; i < k; i++) {
sum += f[i];
}
cout << sum << "\n";
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
import java.awt.*;
public class E1
{
BufferedReader in;
PrintWriter ob;
StringTokenizer st;
public static void main(String[] args) throws IOException {
new E1().run();
}
void run() throws IOException {
in=new BufferedReader(new InputStreamReader(System.in));
ob=new PrintWriter(System.out);
solve();
ob.flush();
}
void solve() throws IOException {
int n = ni() , k = ni();
int a[][] = new int[n][3];
ArrayList<Integer> both = new ArrayList<>();
ArrayList<Integer> cntA = new ArrayList<>();
ArrayList<Integer> cntB = new ArrayList<>();
for(int i=0 ; i<n ; i++) {
a[i][0] = ni();
a[i][1] = ni();
a[i][2] = ni();
if( a[i][1] == 1 && a[i][2] == 1 )
both.add(a[i][0]);
else if( a[i][1] == 1 )
cntA.add(a[i][0]);
else if( a[i][2] == 1 )
cntB.add(a[i][0]);
}
if( isNotValid( a , k ) ) {
ob.println("-1");
return;
}
Collections.sort(both);
Collections.sort(cntA);
Collections.sort(cntB);
int[] cboth = new int[both.size()+1];
for (int i = 0 ; i<both.size() ; i++ ) {
cboth[i+1] = cboth[i] + both.get(i);
}
int preA[] = new int[cntA.size()+1];
for (int i = 0 ; i < cntA.size() ; i++ ) {
preA[i+1] = preA[i] + cntA.get(i);
}
int preB[] = new int[cntB.size()+1];
for (int i = 0 ; i < cntB.size() ; i++ ) {
preB[i+1] = preB[i] + cntB.get(i);
}
long ans = Long.MAX_VALUE;
for(int i=0 ; i<cboth.length ; i++) {
int need = Math.max( 0 , k - i );
if( preA.length>need && preB.length>need )
ans = Math.min( ans , cboth[i] + preA[need] + preB[need] );
}
ob.println(ans);
}
boolean isNotValid(int a[][] , int k) {
int cntA = 0 , cntB = 0;
for(int i=0 ; i<a.length ; i++) {
if( a[i][1] == 1 )
++cntA;
if( a[i][2] == 1 )
++cntB;
}
return !( cntA>=k && cntB>=k );
}
String ns() throws IOException {
return nextToken();
}
long nl() throws IOException {
return Long.parseLong(nextToken());
}
int ni() throws IOException {
return Integer.parseInt(nextToken());
}
double nd() throws IOException {
return Double.parseDouble(nextToken());
}
String nextToken() throws IOException {
if(st==null || !st.hasMoreTokens())
st=new StringTokenizer(in.readLine());
return st.nextToken();
}
int[] nia(int start,int b) throws IOException {
int a[]=new int[b];
for(int i=start;i<b;i++)
a[i]=ni();
return a;
}
long[] nla(int start,int n) throws IOException {
long a[]=new long[n];
for (int i=start; i<n ;i++ ) {
a[i]=nl();
}
return a;
}
public void tr(Object... o) {
System.out.println(Arrays.deepToString(o));
}
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python2
|
from __future__ import division, print_function
def main():
# Template 1.0
import sys, re, math
from collections import deque, defaultdict, Counter, OrderedDict
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, floor
from heapq import heappush, heappop, heapify, nlargest, nsmallest
def STR():
return list(input())
def INT():
return int(input())
def MAP():
return map(int, input().split())
def LIST():
return list(map(int, input().split()))
def list2d(a, b, c):
return [[c] * b for i in range(a)]
def sortListWithIndex(listOfTuples, idx):
return (sorted(listOfTuples, key=lambda x: x[idx]))
def sortDictWithVal(passedDic):
temp = sorted(passedDic.items(), key=lambda kv: (kv[1], kv[0]))
toret = {}
for tup in temp:
toret[tup[0]] = tup[1]
return toret
def sortDictWithKey(passedDic):
return dict(OrderedDict(sorted(passedDic.items())))
INF = float('inf')
mod = 10 ** 9 + 7
n,k = MAP()
a = []
Alen = 0
b = []
Blen = 0
both = []
bothLen = 0
for _ in range(n):
time, A, B = MAP()
if(A==1 and B==1):
both.append(time)
bothLen+=1
elif(A==1):
a.append(time)
Alen+=1
elif(B==1):
b.append(time)
Blen+=1
if((len(both) + len(a)<k) or (len(both) + len(b)<k)):
print(-1)
else:
a.sort()
b.sort()
both.sort()
I,J,K = 0,0,0
ans = 0
while(k!=0):
if(K<bothLen):
temp = both[K]
else:
temp = INF
if(I<Alen):
temp1 = a[I]
else:
temp1 =INF
if(J<Blen):
temp2 = b[J]
else:
temp2 = INF
if(temp<=temp1+temp2):
ans+=temp
K+=1
else:
ans+=temp1+temp2
I+=1
J+=1
k-=1
print(ans)
######## Python 2 and 3 footer by Pajenegod and c1729
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.
# So on cf, use PyPy2 for best string performance.
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
import os, sys
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO, self).read()
def readline(self):
while self.newlines == 0:
s = self._fill();
self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda: self.buffer.read().decode('ascii')
self.readline = lambda: self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
# Cout implemented in Python
import sys
class ostream:
def __lshift__(self, a):
sys.stdout.write(str(a))
return self
cout = ostream()
endl = '\n'
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero=0):
conv = ord if py2 else lambda x: x
A = [];
numb = zero;
sign = 1;
i = 0;
s = sys.stdin.buffer.read()
try:
while True:
if s[i] >= b'0'[0]:
numb = 10 * numb + conv(s[i]) - 48
elif s[i] == b'-'[0]:
sign = -1
elif s[i] != b'\r'[0]:
A.append(sign * numb)
numb = zero;
sign = 1
i += 1
except:
pass
if s and s[-1] >= b'0'[0]:
A.append(sign * numb)
return A
if __name__ == "__main__":
main()
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
from sys import stdin
from collections import deque
# https://codeforces.com/contest/1354/status/D
mod = 10**9 + 7
import sys
import random
# sys.setrecursionlimit(10**6)
from queue import PriorityQueue
from collections import Counter as cc
# def rl():
# return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
import heapq
# input = lambda: sys.stdin.readline().rstrip()
input = lambda : sys.stdin.readline().rstrip()
from sys import stdin, stdout
from heapq import heapify, heappush, heappop
from itertools import permutations
from math import factorial as f
# def ncr(x, y):
# return f(x) // (f(y) * f(x - y))
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
import sys
# input = sys.stdin.readline
# LCA
# def bfs(na):
#
# queue = [na]
# boo[na] = True
# level[na] = 0
#
# while queue!=[]:
#
# z = queue.pop(0)
#
# for i in hash[z]:
#
# if not boo[i]:
#
# queue.append(i)
# level[i] = level[z] + 1
# boo[i] = True
# dp[i][0] = z
#
#
#
# def prec(n):
#
# for i in range(1,20):
#
# for j in range(1,n+1):
# if dp[j][i-1]!=-1:
# dp[j][i] = dp[dp[j][i-1]][i-1]
#
#
# def lca(u,v):
# if level[v] < level[u]:
# u,v = v,u
#
# diff = level[v] - level[u]
#
#
# for i in range(20):
# if ((diff>>i)&1):
# v = dp[v][i]
#
#
# if u == v:
# return u
#
#
# for i in range(19,-1,-1):
# # print(i)
# if dp[u][i] != dp[v][i]:
#
# u = dp[u][i]
# v = dp[v][i]
#
#
# return dp[u][0]
#
# dp = []
#
#
# n = int(input())
#
# for i in range(n + 10):
#
# ka = [-1]*(20)
# dp.append(ka)
# class FenwickTree:
# def __init__(self, x):
# """transform list into BIT"""
# self.bit = x
# for i in range(len(x)):
# j = i | (i + 1)
# if j < len(x):
# x[j] += x[i]
#
# def update(self, idx, x):
# """updates bit[idx] += x"""
# while idx < len(self.bit):
# self.bit[idx] += x
# idx |= idx + 1
#
# def query(self, end):
# """calc sum(bit[:end])"""
# x = 0
# while end:
# x += self.bit[end - 1]
# end &= end - 1
# return x
#
# def find_kth_smallest(self, k):
# """Find largest idx such that sum(bit[:idx]) <= k"""
# idx = -1
# for d in reversed(range(len(self.bit).bit_length())):
# right_idx = idx + (1 << d)
# if right_idx < len(self.bit) and k >= self.bit[right_idx]:
# idx = right_idx
# k -= self.bit[idx]
# return idx + 1
# import sys
# def rs(): return sys.stdin.readline().strip()
# def ri(): return int(sys.stdin.readline())
# def ria(): return list(map(int, sys.stdin.readline().split()))
# def prn(n): sys.stdout.write(str(n))
# def pia(a): sys.stdout.write(' '.join([str(s) for s in a]))
#
#
# import gc, os
#
# ii = 0
# _inp = b''
#
#
# def getchar():
# global ii, _inp
# if ii >= len(_inp):
# _inp = os.read(0, 100000)
# gc.collect()
# ii = 0
# if not _inp:
# return b' '[0]
# ii += 1
# return _inp[ii - 1]
#
#
# def input():
# c = getchar()
# if c == b'-'[0]:
# x = 0
# sign = 1
# else:
# x = c - b'0'[0]
# sign = 0
# c = getchar()
# while c >= b'0'[0]:
# x = 10 * x + c - b'0'[0]
# c = getchar()
# if c == b'\r'[0]:
# getchar()
# return -x if sign else x
# fenwick Tree
# n,q = map(int,input().split())
#
#
# l1 = list(map(int,input().split()))
#
# l2 = list(map(int,input().split()))
#
# bit = [0]*(10**6 + 1)
#
# def update(i,add,bit):
#
# while i>0 and i<len(bit):
#
# bit[i]+=add
# i = i + (i&(-i))
#
#
# def sum(i,bit):
# ans = 0
# while i>0:
#
# ans+=bit[i]
# i = i - (i & ( -i))
#
#
# return ans
#
# def find_smallest(k,bit):
#
# l = 0
# h = len(bit)
# while l<h:
#
# mid = (l+h)//2
# if k <= sum(mid,bit):
# h = mid
# else:
# l = mid + 1
#
#
# return l
#
#
# def insert(x,bit):
# update(x,1,bit)
#
# def delete(x,bit):
# update(x,-1,bit)
# fa = set()
#
# for i in l1:
# insert(i,bit)
#
#
# for i in l2:
# if i>0:
# insert(i,bit)
#
# else:
# z = find_smallest(-i,bit)
#
# delete(z,bit)
#
#
# # print(bit)
# if len(set(bit)) == 1:
# print(0)
# else:
# for i in range(1,n+1):
# z = find_smallest(i,bit)
# if z!=0:
# print(z)
# break
#
# service time problem
# def solve2(s,a,b,hash,z,cnt):
# temp = cnt.copy()
# x,y = hash[a],hash[b]
# i = 0
# j = len(s)-1
#
# while z:
#
# if s[j] - y>=x-s[i]:
# if temp[s[j]]-1 == 0:
# j-=1
# temp[s[j]]-=1
# z-=1
#
#
# else:
# if temp[s[i]]-1 == 0:
# i+=1
#
# temp[s[i]]-=1
# z-=1
#
# return s[i:j+1]
#
#
#
#
#
# def solve1(l,s,posn,z,hash):
#
# ans = []
# for i in l:
# a,b = i
# ka = solve2(s,a,b,posn,z,hash)
# ans.append(ka)
#
# return ans
#
# def consistent(input, window, min_entries, max_entries, tolerance):
#
# l = input
# n = len(l)
# l.sort()
# s = list(set(l))
# s.sort()
#
# if min_entries<=n<=max_entries:
#
# if s[-1] - s[0]<window:
# return True
# hash = defaultdict(int)
# posn = defaultdict(int)
# for i in l:
# hash[i]+=1
#
# z = (tolerance*(n))//100
# poss_window = set()
#
#
# for i in range(len(s)):
# posn[i] = l[i]
# for j in range(i+1,len(s)):
# if s[j]-s[i] == window:
# poss_window.add((s[i],s[j]))
#
# if poss_window!=set():
# print(poss_window)
# ans = solve1(poss_window,s,posn,z,hash)
# print(ans)
#
#
# else:
# pass
#
# else:
# return False
#
#
#
#
# l = list(map(int,input().split()))
#
# min_ent,max_ent = map(int,input().split())
# w = int(input())
# tol = int(input())
# consistent(l, w, min_ent, max_ent, tol)
# t = int(input())
#
# for i in range(t):
#
# n,x = map(int,input().split())
#
# l = list(map(int,input().split()))
#
# e,o = 0,0
#
# for i in l:
# if i%2 == 0:
# e+=1
# else:
# o+=1
#
# if e+o>=x and o!=0:
# z = e+o - x
# if z == 0:
# if o%2 == 0:
# print('No')
# else:
# print('Yes')
# continue
# if o%2 == 0:
# o-=1
# z-=1
# if e>=z:
# print('Yes')
# else:
# z-=e
# o-=z
# if o%2!=0:
# print('Yes')
# else:
# print('No')
#
# else:
#
# if e>=z:
# print('Yes')
# else:
# z-=e
# o-=z
# if o%2!=0:
# print('Yes')
# else:
# print('No')
# else:
# print('No')
#
#
#
#
#
#
#
# def dfs(n):
# boo[n] = True
# dp2[n] = 1
# for i in hash[n]:
# if not boo[i]:
#
# dfs(i)
# dp2[n] += dp2[i]
#
#
# n = int(input())
# x = 0
# l = list(map(int,input().split()))
# for i in range(n):
#
# x = l[i]|x
#
# z = list(bin(x)[2:])
# ha = z.copy()
# k = z.count('1')
# ans = 0
# # print(z)
# cnt = 0
# for i in range(20):
#
#
# maxi = 0
# idx = -1
#
# for j in range(n):
# k = bin(l[j])[2:]
# k = '0'*(len(z) -len(k)) + k
# cnt = 0
# for i in range(len(z)):
# if k[i] == z[i] == '1':
# cnt+=1
#
# maxi = max(maxi,cnt)
# if maxi == cnt:
# idx = j
# if idx!=-1:
#
# k = bin(l[idx])[2:]
# k = '0'*(len(z) -len(k)) + k
#
# for i in range(len(z)):
# if k[i] == z[i] == '1':
# z[i] = '0'
# l[idx] = 0
#
#
# ans = 0
# for i in range(len(z)):
# if z[i] == '0' and ha[i] == '1':
# ans+=1
# flip = 0
# for i in range(ans):
# flip+=2**i
#
#
#
# print(flip)
# def search(k,l,low):
#
# high = len(l)-1
# z = bisect_left(l,k,low,high)
#
# return z
#
#
#
#
#
#
# n,x = map(int,input().split())
#
# l = list(map(int,input().split()))
#
# prefix = [0]
# ha = [0]
# for i in l:
# prefix.append(i + prefix[-1])
# ha.append((i*(i+1))//2 + ha[-1])
# fin = 0
# print(prefix)
# for i in range(n):
# ans = 0
# if l[i]<x:
#
#
# if prefix[-1]-prefix[i]>=x:
#
# z = search(x+prefix[i],prefix,i+1)
# print(z)
# z+=i+1
# k1 = x-(prefix[z-1]-prefix[i])
# ans+=ha[z-1]
# ans+=(k1*(k1+1))//2
#
#
# else:
# z1 = x - (prefix[-1]-prefix[i])
# z = search(z1,prefix,1)
#
# k1 = x-prefix[z-1]
# ans+=ha[z-1]
# ans+=(k1*(k1+1))//2
#
#
#
#
# elif l[i]>x:
# z1 = ((l[i])*(l[i]+1))//2
# z2 = ((l[i]-x)*(l[i]-x+1))//2
# ans+=z1-z2
# else:
# ans+=(x*(x+1))//2
# t = int(input())
#
# for _ in range(t):
#
# s = list(input())
# ans = [s[0]]
#
# for i in range(1,len(s)-1,2):
# ans.append(s[i])
#
# ans.append(s[-1])
# print(''.join(ans))
#
#
#
# t = int(input())
#
# for _ in range(t):
#
# n = int(input())
# l = list(map(int,input().split()))
# ans = 0
# for i in range(n):
# if l[i]%2!=i%2:
# ans+=1
#
# if ans%2 == 0:
# print(ans//2)
# else:
# print(-1)
# t = int(input())
#
# for _ in range(t):
#
# n,k = map(int,input().split())
# s = input()
#
#
# ans = 0
# ba = []
# for i in range(n):
# if s[i] == '1':
# ba.append(i)
# if ba == []:
# for i in range(0,n,k+1):
# ans+=1
#
# print(ans)
# continue
#
# i = s.index('1')
# c = 0
# for x in range(i-1,-1,-1):
# if c == k:
# c = 0
# ans+=1
# else:
# c+=1
#
#
# while i<len(s):
# count = 0
# dis = 0
# while s[i] == '0':
# dis+=1
# if dis == k:
# dis = 0
# count+=1
# if dis<k and s[i] == '1':
# count-=1
# break
# i+=1
# if i == n:
# break
# i+=1
#
# print(ans)
#
#
#
#
# q1
# def poss1(x):
# cnt = 1
# mini = inf
# for i in l:
# if cnt%2 != 0:
# cnt+=1
# else:
# if i<=x:
# cnt+=1
# mini = min()
# if cnt == k:
# return
#
# return -1
#
#
#
#
# def poss2(x):
# cnt = 1
# for i in l:
# if cnt%2 == 0:
# cnt+=1
# else:
# if i<=x:
# cnt+=1
# if cnt == k:
# return
#
# return -1
#
#
#
# n,k = map(int,input().split())
# l = list(map(int,input().split()))
#
# z1 = min(l)
# z2 = max(l)
# t = int(input())
#
# for _ in range(t):
#
# n,k = map(int,input().split())
# l = list(map(int,input().split()))
#
# for i in range(n):
# z = l[i]%k
#
# if z!=0:
#
# l[i] = k-z
# else:
# l[i] = 0
#
# l.sort()
# count = 0
# ans = 0
# x = 0
# print(l)
# for i in range(n):
#
# z = l[i]
# if z == 0:
# continue
# if x>z:
# z2 = ceil((x-l[i])/k)
# z1 = l[i] + k*z2
# # print(x,z1)
# ans += z1-x + 1
# x = z1+1
# elif z == x:
# ans+=1
# x+=1
# else:
#
# ans += z-x + 1
# x = z+1
# # print(i,x)
# print(ans)
#
#
#
# print(ans)
n,k = map(int,input().split())
ha = k
la1 = []
la2 = []
la3 = []
for i in range(n):
t,a,b = map(int,input().split())
if a == 1 and b == 0:
la1.append(t)
elif a == 0 and b == 1:
la2.append(t)
elif a == 1 and b == 1:
la3.append(t)
la1.sort()
la2.sort()
la3.sort()
ans = 0
count1 = 0
count2 = 0
i,j,k = 0,0,0
z = max(len(la1),len(la2),len(la3))
# print(la1,la2,la3)
while True:
if i>=z and j>=z and k>=z:
break
if count1 >= ha and count2>=ha:
break
t1,t2,t3 = 10**18,10**18,10**18
try:
t1 = la1[i]
except:
pass
try:
t2 = la2[j]
except:
pass
try:
t3 = la3[k]
except:
pass
if count1>=ha and count2>=ha:
break
elif count1<ha and count2<ha:
if t1+t2<=t3:
count1+=1
count2+=1
ans+=t1+t2
i+=1
j+=1
else:
count1+=1
count2+=1
ans+=t3
k+=1
elif count1<ha and count2>=ha:
if t1<=t3:
i+=1
ans+=t1
else:
k+=1
ans+=t3
elif count2<ha and count1>=ha:
if t2<=t3:
j+=1
ans+=t2
else:
k+=1
ans+=t3
if ans>=10**18:
print(-1)
else:
print(ans)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.Collections;
import java.util.ArrayList;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author fakhoury
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
CReadingBooksEasyVersion solver = new CReadingBooksEasyVersion();
solver.solve(1, in, out);
out.close();
}
static class CReadingBooksEasyVersion {
public void solve(int testNumber, InputReader in, OutputWriter out) {
int n = in.nextInt();
int k = in.nextInt();
ArrayList<Integer> both = new ArrayList<>();
ArrayList<Integer> a = new ArrayList<>();
ArrayList<Integer> b = new ArrayList<>();
a.add(0);
b.add(0);
both.add(0);
for (int i = 0; i < n; i++) {
int time = in.nextInt();
boolean a_like = in.nextInt() == 1;
boolean b_like = in.nextInt() == 1;
if (a_like && b_like) {
both.add(time);
} else if (a_like) {
a.add(time);
} else if (b_like) {
b.add(time);
}
}
Collections.sort(both);
Collections.sort(a);
Collections.sort(b);
for (int i = 1; i < both.size(); i++) {
both.set(i, both.get(i) + both.get(i - 1));
}
for (int i = 1; i < a.size(); i++) {
a.set(i, a.get(i) + a.get(i - 1));
}
for (int i = 1; i < b.size(); i++) {
b.set(i, b.get(i) + b.get(i - 1));
}
int answer = Integer.MAX_VALUE;
for (int qt_both = 0; qt_both <= Math.min(k, both.size() - 1); qt_both++) {
int qt_a = k - qt_both;
int qt_b = k - qt_both;
if (qt_a >= a.size() || qt_b >= b.size()) {
continue;
}
answer = Math.min(answer, both.get(qt_both) + a.get(qt_a) + b.get(qt_b));
}
out.println(answer < Integer.MAX_VALUE ? answer : -1);
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void close() {
writer.close();
}
public void println(int i) {
writer.println(i);
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
# sys.setrecursionlimit(5*10**5)
inf = 10**20
mod = 10**9 + 7
def LI(): return list(map(int, input().split()))
def II(): return int(input())
def LS(): return list(input().split())
def S(): return input()
def main():
n, k = LI()
tab = [LI() for _ in range(n)]
both = list()
only_a = list()
only_b = list()
for t, a, b in tab:
if a and b:
both.append(t)
elif a:
only_a.append(t)
elif b:
only_b.append(t)
m = min(len(only_a), len(only_b))
only_b.sort()
only_a.sort()
for i in range(m):
tmp = only_b[i] + only_a[i]
both.append(tmp)
both.sort()
if len(both) < k:
return -1
return sum(both[:k])
print(main())
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
from bisect import bisect_left,bisect_right
import threading
from collections import Counter,defaultdict
arr=[]
def main():
for _ in range(1):
n,k=map(int,input().split())
ar3=[]
ar1=[]
ar2=[]
for i in range(n):
a,b,c=map(int,input().split())
if b==c==1:
ar3.append(a)
else:
if b==1:
ar1.append(a)
if c==1:
ar2.append(a)
t=2*max(0,k-len(ar3))+len(ar3)
if len(ar3)+len(ar1)<k or len(ar3)+len(ar2)<k:
print(-1)
else:
ar3.sort()
ar1.sort()
ar2.sort()
# print(ar1,ar2,ar3)
pt1=pt2=0
i1=0
i2=0
i3=0
ans=0
m_=0
while i1<len(ar1) and i2<len(ar2) and i3<len(ar3):
if ar1[i1] + ar2[i2] <ar3[i3]:
ans=ans+ar1[i1]+ar2[i2]
i1+=1
i2+=1
else:
ans=ans+ar3[i3]
i3+=1
pt1+=1
pt2+=1
if pt1==pt2==k:
break
# print(pt1,pt2,i1,i2,i3)
if i2>=len(ar2):
i1,i2=i2,i1
ar1,ar2=ar2,ar1
while pt1<k:
if i1<len(ar1) and i3<len(ar3) and ar3[i3]<ar1[i1]:
ans+=ar3[i3]
i3+=1
pt2+=1
elif i1<len(ar1) :
ans+=ar1[i1]
i1+=1
else:
ans+=ar3[i3]
i3+=1
pt2+=1
pt1+=1
while pt2<k:
if i1<len(ar1) and i3<len(ar3) and ar3[i3]<ar1[i1]:
ans+=ar3[i3]
i3+=1
elif i2<len(ar2) :
ans+=ar2[i2]
i2+=1
else:
ans+=ar3[i3]
i3+=1
pt2+=1
print(ans)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
//============================================================================
/* "There is nothing that can take the pain away.
But eventually you will find a way to live with it.
There will be nightmares.
And every day when you wake up, it will be the first thing you think about.
Until one day, it will be the second thing."
*/
// Author : Murad
// Online Judge: Codeforces.cpp & Atcoder.cpp
// Description : Problem name
//============================================================================
/* Riven && Vladimir && Ekko */
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
int n=in.nextInt(),k=in.nextInt();
PriorityQueue<Integer>alic=new PriorityQueue<>();
PriorityQueue<Integer>bob=new PriorityQueue<>();
PriorityQueue<Integer>both=new PriorityQueue<>();
long ans=0;
for(int i=0;i<n;i++)
{
int t=in.nextInt(),a=in.nextInt(),b=in.nextInt();
if(a==1 && b==0)
alic.add(t);
else if(a==0 && b==1)
bob.add(t);
else if(a== 1 &&b==1)
both.add(t);
}
int cnta=(int)both.size()+(int)alic.size();
int cntb=(int)both.size()+(int)bob.size();
if (cnta < k || cntb < k){
out.println(-1);
}
else{
cnta=0;
cntb=0;
int aidx=0,bidx=0;
while (both.isEmpty()==false){
int total=0;
int fialic=0,fibob=0;
if(alic.isEmpty()==false) {
fialic=alic.peek();
total += fialic;
}
else total+=1000000000;
if(bob.isEmpty()==false){
fibob=bob.peek();
total+=fibob;
}else total+=1000000000;
int mo=both.peek();
if(total<mo){
ans += total;
alic.poll();
bob.poll();
cnta++;
cntb++;
}
else{
ans += mo;
cnta++;
cntb++;
both.poll();
}
if (cnta >= k&&cntb >= k)break;
}
while (!alic.isEmpty() && cnta<k) {
int fia=alic.peek();
ans +=fia;
alic.poll();
cnta++;
}
while (!bob.isEmpty() && cntb < k) {
int fib=bob.peek();
ans +=fib;
bob.poll();
cntb++;
}
out.println(ans);
}
out.flush();
}
static class Pair<C, I extends Number> implements Comparable<Pair<C, Number>> {
long value;
long idx;
Pair(long v, long i)
{
value = v;
idx = i;
}
@Override
public int compareTo(Pair<C, Number> p) {
return (int)(value - p.value);
}
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public long[] readLongArray(int n) {
long[] x = new long[n];
for (int i = 0; i < n; i++) {
x[i] = nextLong();
}
return x;
}
public int[] readIntArray(int n) {
int[] x = new int[n];
for (int i = 0; i < n; i++) {
x[i] = nextInt();
}
return x;
}
}
static class NumberTheory{
public static long gcd(long a,long b){
long c;
while (a != 0) {
c = a;
a = b % a;
b = c;
}
return b;
}
}
//Relatively Prime :- if diffrence between two number is equal to 1
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.List;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.util.Collections;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
MyScanner in = new MyScanner(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskE1 solver = new TaskE1();
solver.solve(1, in, out);
out.close();
}
static class TaskE1 {
public void solve(int testNumber, MyScanner in, PrintWriter out) {
int n, k;
n = in.nextInt();
k = in.nextInt();
List<Integer> alice = new ArrayList<>();
List<Integer> bob = new ArrayList<>();
List<Integer> timesBoth = new ArrayList<>();
for (int i = 0; i < n; i++) {
int t = in.nextInt();
int a = in.nextInt();
int b = in.nextInt();
if (a == b && a == 1) {
timesBoth.add(t);
continue;
}
if (a == 1) {
alice.add(t);
}
if (b == 1) {
bob.add(t);
}
}
if (alice.size() + timesBoth.size() < k || bob.size() + timesBoth.size() < k) {
out.println(-1);
} else {
Collections.sort(alice);
Collections.sort(bob);
int i = 0, j = 0;
while (i < alice.size() && j < bob.size()) {
timesBoth.add(alice.get(i) + bob.get(i));
i++;
j++;
}
Collections.sort(timesBoth);
long ans = 0;
for (int i1 = 0; i1 < k; i1++) {
ans += timesBoth.get(i1);
}
out.println(ans);
}
}
}
static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner(InputStream io) {
br = new BufferedReader(new InputStreamReader(io));
}
public String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
def binary(a,start,end,e):
mid=(start+end)//2
if a[mid]==e:
return mid
elif a[mid]>e:
end=mid-1
else:
start=mid+1
if start<=end:
return binary(a,start,end,e)
else:
return start
n,k=map(int,input().split())
p=[]
al=[]
bo=[]
for j in range(n):
a,b,c=map(int,input().split())
if b==1 and c==1:
p.append(a)
elif b==1 and c==0:
al.append(a)
elif b==0 and c==1:
bo.append(a)
al.sort()
bo.sort()
p.sort()
if p!=[]:
while al!=[] and bo!=[]:
v=al[0]+bo[0]
q=binary(p,0,len(p)-1,v)
p.insert(q,v)
del al[0]
del bo[0]
else:
while al!=[] and bo!=[]:
v=al[0]+bo[0]
p.append(v)
del al[0]
del bo[0]
if len(p)<k:
print(-1)
else:
print(sum(p[:k]))
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;
import java.io.BufferedReader;
import java.io.InputStreamReader;
public class First {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskA solver = new TaskA();
//int a = 1;
int t;
//t = in.nextInt();
t = 1;
while (t > 0) {
//out.print("Case #"+(a++)+": ");
solver.call(in,out);
t--;
}
out.close();
}
static class TaskA {
public void call(InputReader in, PrintWriter out) {
int n, k, t, a, b;
n = in.nextInt();
k = in.nextInt();
LinkedList<Integer> alice = new LinkedList<>();
LinkedList<Integer> bob = new LinkedList<>();
LinkedList<Integer> both = new LinkedList<>();
for (int i = 0; i < n; i++) {
t = in.nextInt();
a = in.nextInt();
b = in.nextInt();
if(a+b==2){
both.add(t);
}
else if(a+b==1){
if(a==1){
alice.add(t);
}
else{
bob.add(t);
}
}
}
Collections.sort(bob);
Collections.sort(both);
Collections.sort(alice);
if(both.size() + Math.min(bob.size(), alice.size())<k){
out.println(-1);
return;
}
long ans = 0;
for (int i = 0; i < k; i++) {
if(both.size()==0){
ans+=bob.get(0) + alice.get(0);
bob.remove(0);
alice.remove(0);
continue;
}
if(bob.size()==0 || alice.size()==0){
ans+= both.get(0);
both.remove(0);
continue;
}
if(both.get(0)< bob.get(0) + alice.get(0)){
ans+= both.get(0);
both.remove(0);
}
else{
ans+=bob.get(0) + alice.get(0);
bob.remove(0);
alice.remove(0);
}
}
out.println(ans);
}
}
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
static class answer implements Comparable<answer>{
int a;
int b;
public answer(int a, int b) {
this.a = a;
this.b = b;
}
@Override
public int compareTo(answer o) {
return this.a - o.a;
}
}
static class answer1 implements Comparable<answer1>{
int a, b, c;
public answer1(int a, int b, int c) {
this.a = a;
this.b = b;
this.c = c;
}
@Override
public int compareTo(answer1 o) {
return this.a - o.a;
}
}
static long gcd(long a, long b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static void sort(long[] a) {
ArrayList<Long> l=new ArrayList<>();
for (long i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static final Random random=new Random();
static void shuffleSort(int[] a) {
int n=a.length;
for (int i=0; i<n; i++) {
int oi=random.nextInt(n), temp=a[oi];
a[oi]=a[i]; a[i]=temp;
}
Arrays.sort(a);
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong(){
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
#pragma GCC optimize("02")
#pragma G++ optimize("03")
using namespace std;
const int N = 2e5 + 5;
int n, k, type1[N], type2[N], type3[N];
int main() {
scanf("%d%d", &n, &k);
int cnt1 = 0, cnt2 = 0, cnt3 = 0;
for (int i = 1; i <= n; i++) {
int val, a, b;
scanf("%d%d%d", &val, &a, &b);
if (!a && !b) continue;
if (a && b) type1[++cnt1] = val;
if (a && !b) type2[++cnt2] = val;
if (!a && b) type3[++cnt3] = val;
}
sort(type1 + 1, type1 + 1 + cnt1);
sort(type2 + 1, type2 + 1 + cnt2);
sort(type3 + 1, type3 + 1 + cnt3);
for (int i = 1; i <= min(cnt2, cnt3); i++)
type1[++cnt1] = type2[i] + type3[i];
sort(type1 + 1, type1 + 1 + cnt1);
if (cnt1 < k)
puts("-1");
else {
int res = 0;
for (int i = 1; i <= k; i++) res += type1[i];
printf("%d\n", res);
}
return 0;
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
# from bisect import bisect_left
# TC = int(input())
# for tc in range(TC):
N, K = map(int, input().split())
T = []
A = []
B = []
O = []
for b in range(N):
v, a, b = map(int, input().split())
if a == 1 and b == 1:
T.append(v)
elif a == 1:
A.append(v)
elif b == 1:
B.append(v)
else:
O.append(v)
A.sort()
B.sort()
S = []
for i in range(min(len(A),len(B))):
S.append(A[i]+B[i])
R = S + T
if len(R) >= K:
R.sort()
print(sum(R[:K]))
else:
print(-1)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
n, k = [int(i) for i in input().split()]
doub = []; a = []; b = []
for i in range(n):
ti, ai, bi = [int(i) for i in input().split()]
if ai and bi:
doub.append(ti)
elif ai:
a.append(ti)
elif bi:
b.append(ti)
doub.sort(); a.sort(); b.sort()
# print(doub, a, b, sep="\n")
time = 0
i = 0; j = 0
while i+j < k:
if (i < len(doub)):
db = doub[i]
else:
db = 100000
if (j < len(a) and j < len(b)):
sing = a[j] + b[j]
else:
sing = 100000
if (db <= sing):
time += db
i += 1
else:
time += sing
j += 1
if (len(doub) + min(len(a), len(b))) >= k:
print(time)
else:
print(-1)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
import sys
n, k = [int(e) for e in input().split(' ')]
books = list()
books_map = dict()
for i in range(n):
t, a, b = [int(e) for e in input().split(' ')]
books.append((t, a, b, i))
books_map[i] = (t, a, b)
books = sorted(books, key = lambda x : x[0] * (1 if x[1] == 1 else 1e7))
alice_pay_time = 0
alice_liked_books = list()
for book in books:
if len(alice_liked_books) >= k:
break
if book[1] == 1 and book[2] == 0:
alice_pay_time += book[0]
alice_liked_books.append(book[3])
books = sorted(books, key = lambda x : x[0] * (1 if x[2] == 1 else 1e7))
bob_pay_time = 0
bob_liked_books = list()
for book in books:
if len(bob_liked_books) >= k:
break
if book[2] == 1 and book[1] == 0:
bob_pay_time += book[0]
bob_liked_books.append(book[3])
books = sorted(books, key = lambda x : x[0] * (1 if x[1] == 1 and x[2] == 1 else 1e7))
pay_time = alice_pay_time + bob_pay_time
alice_liked_book_count = len(alice_liked_books)
bob_liked_book_count = len(bob_liked_books)
for book in books:
if book[1] == 1 and book[2] == 1:
if alice_liked_book_count < k or bob_liked_book_count < k:
pay_time += book[0]
alice_liked_book_count += 1
bob_liked_book_count += 1
if alice_liked_book_count > k:
pay_time -= books_map[alice_liked_books[-1]][0]
alice_liked_books.pop()
alice_liked_book_count = k
if bob_liked_book_count > k:
pay_time -= books_map[bob_liked_books[-1]][0]
bob_liked_books.pop()
bob_liked_book_count = k
elif len(alice_liked_books) == 0 or len(bob_liked_books) == 0:
break
elif book[0] < books_map[alice_liked_books[-1]][0] + books_map[bob_liked_books[-1]][0]:
pay_time -= books_map[alice_liked_books[-1]][0]
pay_time -= books_map[bob_liked_books[-1]][0]
pay_time += book[0]
alice_liked_books.pop()
bob_liked_books.pop()
if alice_liked_book_count >= k and bob_liked_book_count >= k:
print(pay_time)
else:
print(-1)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
// No sorceries shall prevail. //
import java.util.*;
import java.io.*;
public class InVoker {
//Variables
static long mod = 1000000007;
static long mod2 = 998244353;
static FastReader inp= new FastReader();
static PrintWriter out= new PrintWriter(System.out);
public static void main(String args[]) {
InVoker g=new InVoker();
g.main();
out.close();
}
//Main
void main() {
int n=inp.nextInt();
int k=inp.nextInt();
int both[]=new int[n+1];
int left[]=new int[n+1];
int right[]=new int[n+1];
long gg=Long.MAX_VALUE;
int c=1,d=1,e=1;
for(int i=0;i<n;i++) {
int t=inp.nextInt();
int x=inp.nextInt();
int y=inp.nextInt();
if(x==1 &&y==1) {
both[c++]=t;
}else if(x==1) {
left[d++]=t;
}else if(y==1) {
right[e++]=t;
}
}
Arrays.sort(both,1,c);
Arrays.sort(left,1,d);
Arrays.sort(right,1,e);
//out.print(both[0]+" ");
for(int i=1;i<c;i++) {
both[i]=both[i-1]+both[i];
//out.print(both[i]+" ");
}
//out.println();
//out.print(left[0]+" ");
for(int i=1;i<d;i++) {
left[i]=left[i]+left[i-1];
//out.print(left[i]+" ");
}
//out.println();
//out.print(right[0]+" ");
for(int i=1;i<e;i++) {
right[i]=right[i]+right[i-1];
//out.print(right[i]+" ");
}
//out.println();
//out.println(c+" "+d+" "+e);
for(int i=0;i<c && i<=k;i++) {
int req=k-i;
if(req<d && req<e) {
gg=Math.min(gg, both[i]+left[req]+right[req]);
}
}
out.println(gg==Long.MAX_VALUE?-1:gg);
}
/*********************************************************************************************************************************************************************************************************
* ti;. .:,:i: :,;;itt;. fDDEDDEEEEEEKEEEEEKEEEEEEEEEEEEEEEEE###WKKKKKKKKKKKKKKKKKKKKWWWWWWWWWWWWWWWWWWW#WWWWWKKKKKEE :,:. f::::. .,ijLGDDDDDDDEEEEEEE*
*ti;. .:,:i: .:,;itt;: GLDEEGEEEEEEEEEEEEEEEEEEDEEEEEEEEEEE#W#WEKKKKKKKKKKKKKKKKKKKKKKKWWWWWWWWWWWWWWWWWWWWWWKKKKKKG. .::. f:,...,ijLGDDDDDDDDEEEEEE *
*ti;. .:,:i: :,;;iti, :fDDEEEEEEEEEEEEEEEKEEEEDEEEEEEEEEEEW##WEEEKKKKKKKKKKKKKKKKKKKKKWWWWWWWWWWWWWWWWWWWWWWWKKKKKKEG .::. .f,::,ijLGDDDDDDDDEEEEEE *
*ti;. .:,:i: .,,;iti;. LDDEEEEEEEEEEKEEEEWEEEDDEEEEEEEEEEE#WWWEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKWWWWWWWWWWWWWWWWWWWKKKKKEDj .::. .:L;;ijfGDDDDDDDDDEEEEE *
*ti;. .:,:i: .:,;;iii:LLDEEEEEEEEEEEKEEEEEEEEDEEEEEEEEEEEW#WWEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKWWKWWWWWWWWWWWWWWKKKKKKKEL .::. .:;LijLGGDDDDDDDDEEEEE *
*ti;. .:,:;: :,;;ittfDEEEEEEEEEEEEEEEEKEEEGEEEEEEEEEEEKWWWEEEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKKWWWWWWWWWWWWWWWKKKKKKKELj .::. :,;jffGGDDDDDDDDDEEEE *
*ti;. .:,:i: .,;;tGGDEEEEEEEEEEEKEEEKEEEDEEEEEEEEEEEEWWWEEEEEEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKKWWWWWWKWWWWWWKKKKKKKEEL .::. .:;itDGGDDDDDDDDDEEEE *
*ti;. .:::;: :;ifDEEEEEEEEEEEEKEEEKEEEEEEEEEEEEEEEWWWEEEEEEEEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKW#WWWKKKKKKKKEEf .::. :,itfGEDDDDDDDDDDDEE *
*ti;. .:::;: :GGEEEEEEEEEEEKEKEEKEEEEEEEEEEEEEEEEWWEEEEEEEEEEEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKW#WKKKKKKKKKEEDG .::. .,;jfLGKDLDDDDDDEEDD *
*ti;. .:::;: fDEEEEEEKKKKKKKKKEKEEEEEEEEEEEEEEE#WEEEEEEEEEEEEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKW#KKKKKKKKKKEEf .:::. .,;tfLGDEDDDDDDDDEEE *
*ti;. :::;: fDEEEEEEKKKKKKKKKKWKEEEEEEEEEEEEEEEWKEEEEEEEEEEEEEEEEEEEEKEKKKKKKKKKKKKKKKKKKKKKKKKKKKKW##KKKKKKKKKEEft :::. .,;tfLGDDDKDDDDDDDDD *
*ti;. .::;: fDEEEEEEKKKKKKKWKKKKKEEEEEEEEEEEEE#WEEWEEEEEDEEDEEEEEEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKKKKW#WKKKKKKKKEEGG :,:. .,;tfLGGDDDKDDDDDDDD *
*ti;. .:.;: tGDEEEEKKKKKKKKKKKKKKKKKEEEEEEEEEEEWEEKWEEEEEEEDEEEEEEEEEEEEEEKEKKKKKKKKKKKKKKKKKKKKKKKKKKWWWKKKKKKKEEDf :::. .,;tfLGGDDDDEDDDDDDD *
*ti;. .::;: fDEEEEEKKKKKKKKKKKWKKKKKKKKEEEEEEEWWEEWEEEEEEEEEEEEEEEEEEEEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKW##KKKKKKKEEEft.::. .,;tfLGGDDDDDDEDDDDD *
*ti;. .:.;: tGDEEEKKKKKKKKKKKKKKKKKKKKKKEKEEEEE#EEWWEEEEEEEEEEEEEEEEEEEEEEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKW#WKKKKKKEEEGD:::. .,;tfLGGDDDDDDDEDDDD *
*ti;. .:.,. LDEEEEKKKKKKKKKKWKWKKKKKKKKKKKKEEEKWEKW#EEEEEEEEEEEEEEEEKEEEEEEEEEEEEEEEKKKKKKKKKKKKKKKKKKKKW##KKKKKKEEEEf,,:. .,;tfLGGDDDDDDDDEDDD *
*ti;. ..:.,. LGDEEEEKKKKKKKKKKWKKKKKKKKKKKKKKKKKWEEW#WEEEEEEEEEEEEEEEKEEEEEEEEEEEEEEEEEEEKEKKKKKKKKKKKKKKKK##KKKKKEEEEEfi;,. .,;tfLGGDDDDDDDDDKDD *
*tt;. .:.,: jDEEEEKKKKKKKKKKWWKKKKKKKKKKKKKKKKKWKE#WWEEEEEEEEEEEEEEWEEEEEEEEEEEEEEEEEEEEEEEKKKKKKKKKKKKKKKKWWKKKKEEEEDfG;,: .,;tfLGGDDDDDDDDDDKD *
*tii,. .:.,. tGDEEEEKKKKKKKKKKWWWKKKKKKKKKKKKKKKWKKWWWKEEEEEEEEEEEEEKEEEEEEEEEEEEEEEEEEEEEEEEEEEEEKKKKKKKKKKKW#KKKKEEEEDGGi;,. .,;tfLGGDDDDDDDDDDDE *
*ti;;,:. .:.,: fDEEEEKKKKKKKKKKKWKKKKKKKKKKKKKKKKKWEK#WWKEEEEEEEEEEEEDEEEEEEEEEEEEEEGEEEEEEEEEEEEEEEEEEEKKKKKKKWWKKEEEEEEDDf;;;,. .,;tfLGGDDDDDDDDDDDD *
*tii;,,:.. ...,. ;LEEEEEKKKKKKWKKKKWKKKKKKKKKKKKKKKKKEKKW#WEEEEEEEEEEEEEjEEEEEEEEEKEEEEGEEEEEEEEEKEEEEEEEEEEEEEEEEE#WKEEEEEEDDf;;;;,: .,itfLGGDDDDDDDDDDDD *
*ti;,,,,,:. ...,. LDEEEEKKKKKKKKKKKWWWKKKKKKKKKKKKKKKWKK#W#WEEEEEEEEEEEDDLEEEEEEEEEWEEEEDEEEEEEEEEKEEEEEEEEEEEEEEEEEWWEEEEEEEDDfj,,,,,:. .,itfGGGDDDDDDDDDDDD *
*tii,,,,::::. ...,: .fDEEEEKKKKKKWKKKKWWWKKKKKKKKKKKKKKKEKKW#WWEEEEEEEEEEEKiKEEKEEEEEEWEEEEDEEEEEEEEEEEEEEEEEEEEEEEEEEEWWEEEEEEEDDLD:::,,,:. .,ijfGGGDDDDDDDDDDDD *
*ti;:::::::::.. .:.,: LDEEEEKKKKKKKWKKKKWWKKKKKKKKKKKKKKKKtKKWWWWKEEEEEEEEEDiiDEEEEEEEEWWEEEEEEDEEEEEEEEEEEEEEEEEEEEEEEEEEWKEEEEEDDDGL:. .:,,,: .,ijLGGGDDDDDDDDDDDD *
*tt;. .::::::::.. ...,: :fDEEEKKKKKKKKKKKKWW#KKKKKKKKKKKKKKKKfKKWWWWKEEEEEEEEDti,DEKEEEEEEWWEEEDEEEEEEEEEKEEEEEEEEEEEEEDEEEEE#WEEEEEGGDGf:. .:,;,:. .,ijLGGDDDDDDDDDDDDD *
*tt;. .:::::::.. ...,: GDEEEKKKKKKKKWKKKKWWWKKKWKKKKKKKWWWKDEKLWWWWKKEEEEEEDEi,LDEEEEEEEEWWEEEEEEEEEEEEEEEEEEEEEEEEEDEDEEEEEW#EEEEDDDDGf,. :,,,:...,ijLGGGDDDDDDDDDDDD *
*tt;. .....::::.. ...,: fDEEEKKKKKKKKWKKKKWWWWKKKWKKKKKKKKKKfWKiWWW#KKEEEEEEEi;.EDfEEEDEEiWWEEEEEEEEEEEEDGKEEEEEEEEEEDEEEEEEEWWEEEEDDDGGLi. .,;,:::,ijLGGGDDDDDDDDDDDD *
*tt;. ....:::::. ...,. iDEEEEKKKKKKKKWKKWKWWWWWKKWWWKKKKKKKKtWKt#WWWKKEEEEEDji..DDKDDEDEGiWKEEEEEEEEEEDDEjEEEEEEEEEEEDEEEEEEEKWKEEDDDDGGff. .:,;,,;ijLGGGDDDDDDDDDDDD *
*tt;. ....::::.. .:.,: .LDEEEKKKKKKKKKKKKWWWWKWWWWWWWWWWKKKKWtKKiDWWWKKKEEEEKi:..DEDDDDDDiiWKEEEEEEEEEEDDEijDEEEEEKEEEEEEEEEEEEWWEEGDDDGGLG. .:,;;iijLGGGDDDDDDDDDDDD *
*tt;. .....:::.. ...,. .fEEEEKKKKKKKKWKKKKWWWWWWWWWWWWWWKWKKKiKDiLWWWWKEEEEEi,..fD:DDDDDti;WEEEEEEEEEEKDDi:iDDEEEEWEEEEEEEEEEEE#WEEGDDDDGGG. :,iitjLGGGDDDDDDDDDDDD *
*tti. .....:::.. ...,. GDEEEKKKKKKKKKWKKKWWW#WWWWWWWWWWWKWKKjiEjitWWWKKWEEEDi...DDLDDDDji;;WEEEEEEEEEEEDEj.iDDEEEEWEEEEEEEEEEEEWWEEDDDDDDGf. .,;tjfLGGDDDDDDDDDDDD *
*tti. ....::::.. ...,. fEEEKKKKKKKKKKKKKKKW#WWWWWWWWWWWWWWWWtiEiiiWWWKKEWKEi....D.EDDDEi;.fWEEEEEEEEEEDDfL.;EDDEEEWEEEEEEEEEEEEWWEEEDDDDDGf. :;ijfLGGDDDDDDDDDDDD *
*tti. ....::::.. ...,. LDEEEKKKKKKKKKKKKKKWWWWWWWWWWWWWWWW####WKiiiWWWKKKEEK,...:E:DDDEii..GWEEEEEEEEDWDDiL.,KDDEEEWEEEEEEEEEEEEWWKEEDDDDDGf: .,itfLGGDDDDDDDDDDDD *
*tti. .....:::.. ...,. fDEEEKKKKKKKKKWKKKKWWWWWWWWWWWWW########WLiiWWWKKKEEjD...G,DDDDi;...EWEEEEEEEEDKDEii..LDDEEEWEEEEEEEEEEEEWWWEEDDDDDGfi .,;tfLGGGDDDDDDDDDDD *
*tti. .....:::.. ...,. iGEEEKKKKKKKKKKWKKKKWWWWWWWWWWWW###########KiWWWKKEEE,.D..D.DDDii:...KKEEEEEEEEEDDj:...tEDEEEWEEEEEEEEEEEEWWWEEEDDDDDLL .,;tjLLGGDDDDDDDDDDD *
*tti. ....::::......:. LEEEKKKKKKKKKKWWKKKWWW#KWWWWWWWW#####W####W##KWWKKEEL..:D.jjDDi;,....KKEEEEEEEDfDDi...:iKDEEEWKEEEEEEEEEEEWWWEEEEDDDDLG .,;tjLLGGDDDDDDDDDDD *
*tti. ...::::::..,. :GEEEKKKKKKKKKKKKWWWWW##WWWWWWWWW##WKWK#W#W####WWKEEK.....G.DDti,.....KKEEEEEEDWGDf.,...iKDEEEWWEEEEEEEEEEEW#WEEEEEDDDGL .,;tjLLGGDDDDDDDDDDD *
*tti. ....::::::,. GDEEKKKKKKKKKKKKKWWWW###WWWWWWWWWW#WWWK###W#####WKEKK.....jDDL;;......KKEEEEEEEEEDi.f...;KDEEEWWEEEEEEEEEEEWWWWEEEEEDDGf .,;tjLLGGDDDDDDDDDDD *
*tti. ....:::,,. .LEEEKKKKKWKKKKKWWWWWW###WWWWWWWWWW#WWKW#WW##W#WWWKEKD:....:DD:;......;KEEEEEEEKiDD..f...,KKEEEWWEEEEEEEEEEEWWWWEEEEEDDDf .:;tjLLGGGDDDDDDDDDD *
*tti. ...::,,,:. GDEEKKKKKKKKKKKKWWWWWWW#WWWWWWWWWWW#KjKWWWWWWWWWWWWEK.j,..;fD.;.......fKEEEEEDKG:Di..,....DKEEEWWEEEEEEKEKKKWWWWEEEEEEDDf .:;tjLLGGDDDDDDDDDDD *
*jti. ...::,,,,:. .fEEEKKKKKWKKKKKKWWWWWWW#WWWWWWWWWWK#KKKWWWWWWWWWWWWWK..f:.:G.,:.......EKEEEEEKK;:E:.......fKEEEWWKEKEKKKKKKKW#WWEEEEEEDDf: .,;tfLLGGDDDDDDDDDDD *
*tti. ...:,,,;;,: iDEEKKKKKWKKKKKKKWWWWWWW#WWWWWWWWWWK#WDKWWKKWWWWWWWWWE..;G:G..,........KKEEEEEKi.Gi..:.....tKEEKWWWKKKKKKKKKKW##WKEEEEEEDfi .,;tfLLGGGDDDDDDDDDD *
*tti. ....::,,;;;,LEEKKKKKKWKKKKKWWWWWWW###WWWWWWWWWWKWWDKWEEEWKKWWWWWKKj.:LG..;.........EKEEEEKG;.G...;.....;KKEKWWWKKKKKKKKKKW##WWKEEEEEDfL .,;tfLGGGDDDDDDDDDDD *
*jti. ...::::,;ijDEEKKKKKWKKKKKKWKWWWWW##WWWWWWWWWWWKK#KKGDGDWEEWKKWKKGE,.i;.:.........:EKEEEKE;.:L...j.....,KWEKWWWKKKKKKKKKK####WKKEEEEDLG .,;tfLGGGGDDDDDDDDDD *
*jti. ...:...,,;GEEKKKKKWWKKKKKWWWWWWWW###WWWWWWWWWKKKWWKiLGGEDEDEKGKKiEG..;...........jKEEEKK;:.G....,.....:KKEWWWWKKKKKKKWKK####WKKKKEEEGL .,;tfLGGGGDDDDDDDDDD *
*jti. ...:. .:,GEEKKKKKWKKKKKWWWWWWWW####WWWWWWWWWKKKWWKii;fDLGDK: EEi:E:.............EKEEKK;;..L...........KKKWWWWKKKKKKKWKK####WKKKWKEEDf .,;tfGGGGDDDDDDDDDDD *
*jti. ...:. ,EEKKKKKWWKKKKKWWWWWWWWW###WWWWWWWWKKKKfWWLt;i,. fi EG..D:.............EKEKK;;..t....:.......KWKWWWWKKKKKKKWKK####WKKKWEEEDf:. .,;tfGGGGDDDDDDDDDDD *
*jti. ...:. GEEKKKKKWKKKKKWWWWWWWWW####WWWWWWWKKKKKt;KKEfff .;t.................KKKKi;:..GtGGfG.......KWWWWWWKKKKKKKWKK###WWWKKKKEEEf,,: .,;tfGGGGDDDDDDDDDDD *
*jti. ...:. GEKKKKKWWKKKKKWWWWWWWWWW##WWWWWWWKKKKKKt;EiKKKK, ...t................jEKKG;;..,.....,LGi....KWWWWWWKKKKKKWKKKW####WKKKKKEEL,,,:. .,;tfGGGDDDDDDDDDDDD *
*jti. ...:. .GEEKKKKKWKKKKKWWWWWWWWWW###WWWWWWWKKKKKKtiE::tGG........................EEEj;;...,.........:D..DKWWWWWWKKKKK#KKW###W#WKKKKKEEfj:,,,:. .,;tfGGGDDDDDDDDDDDD *
*jti. ...:. DEKKKKKWWKKKKKWWWWWWWWW####WWWWWWWKKKKKKiiE:::.::.......................EEi;;...j.....f......:iDKWWWWWWKKKKK#WW######WKKKKKEELG :,,,,:. .,;tfGGGDDDDDDDDDDDD *
*jti. ...:. fEEKKKKWWKKKKWWWWWWWWWWW###WWWWWWWWKKKKKK;tE::..........................DD;.;,.::......;........EWWWWWWWKKKKW#WW#####WWKKKWKKELG .:,,,:::,;tfGGGDDDDDDDDDDDD *
*jti. ...:. .DEKEKKKWWKKKKWWWWWWWWWWW###WWWWWWWWKKKKKE,iD::..........................D..,;.,;tLffi...........DWDWWWW#KKKWWWWW#####W#KKKWKEEGL .:,;,,,;tfGGGDDDDDDDDDDDD *
*jti. ...:. ;EEKKKKWWKKKKKWWWWWW#WWWW####WWWWWWKKKKKEL:iD:..........................j ..;..;;:.....i,........DKtWWWWWKKWWWWWW#####WWWKKKEKEDf .:,;;;itfGGGDDDDDDDDDDDD *
*jti. ...:. DEKKKKKWWKKKKWWWWWWW#WWWW####WWWWWWKKKKKEj:iG...............................:....................GKiWWWWWKKWW#WWW######WWKKKKKEEf .,;iitfGGGDDDDDDDDDDDD *
*jti. ...:.:EKKKKKWWKKKKKWWWWWWW#WWW#####WWWWWKWKKKKEi:if:.................................iEKEKKKKKKDj......DKiWWWWWKWK##WW#######WWKKK:KEEL .:;itfGGGDDDDDDDDDDDD *
*jji. ...:,DEEKKKWWWKWKKWWWWWWWWWWWW#####WWWWWWWKKKKEi:it..................................j. KKKKKKKKKKKf..DKiWWWWWKWW##WW#######WWKKK,KEEf .,;tfGGGDDDDDDDDDDDD *
*jji. ..L:iDEEKKKWWKKKKKWWWWWWWWWWWW#####WWWWWKWKKKKKi.i;.................................. . KKKWWWWWWWWK..DGiWWWWWKK##WWW#####W#WWKKKjEKEL, .:;tfGGGDDDDDDDDDDDD *
*jji. .f:::EEEKKKWWWKKKKKWWWWWWWWWWWW#####WWWWWKWKKKKK;.i,.................................:: KKEKWWWWWWfWK..EiiWWWWWKWW#WW##########KKKD,KELj .:;tfGGDDDDDDDDDDDDD *
*jji. .t::::,DEEKKKWWKKKKWWWWWWWWW#WWWW#####WWWWKKWKKKEK;.i:.................................GDDEEEKKKWWWWWtWWD.E;iWWWWWW###WW#########WWKKK.EEDG .:;tfGGGDDDDDDDDDDDD *
*jji. . j..::::EKEKKKWWWKKKKWWWWWWWWW#WWW######WWWWKKWKKKEK;.t:.................................ELLEDDEEEWWWWEtWK,.KiiWWWWWW###W##########WWKKK:EEEG .;tjfLLGDDDDDDDDDDDDDDD *
*jji. i.::::::,EEEKKWWWKKKKKWWWWWWWWW#WWW#####WWWWWKWKKKKEE,.t..................................DfiEGDDDEEKKKttKWG.KiiWWWWW##WWW##########WWKKK:fEEL ,fGGGDDDDDDDDEEEDDDDDDDDDD *
*jji. .;:..:::::DEEEKKWWWKKKKKWWWWWWWWW#WWWW####WWWWWWWKKKKED,.t..................................ifjDDGGEGDKK.ttKKE.DiWWWWW###WW##########WWWKKK:.KELiLGGGGDDDDDDDDDDDDEEEDDDDDDD *
*jji. i.:.::::::,KEEKKWWWKKKKKKWWWWWWWWW#WWWW####WWWWWWWKKKKEL:.j..................................GGf,;ifLLED .iiKKi:fWWWWWW##W#W##########WWWKKK:.KKLGGDDDDDDDDDDDDDDDDEDDEEDDDDD *
*jji. .j:.::::::::EEEKKKWWWKKKKKKWWWWWWWW##WWW#####WWWWWWWKKKKKf:.f..................................:EEfftf .,. ;iE,..jWWWWWWW###W############WWKKK,:KKGDDDDDDDDDDDDDDDDDDDDDDDEDDDD *
*jji. .:.::::::::,,EEEKKWWWKKKKKKKWWWWWWWW##WWW#####WWWWWWWKKKKKt..G....................................EEELL; .j....tKWWWWWWW################WWWKKtfGKGEDDDDDDDDDDDDDDDDDDDDDDDEEDD *
*jji. :...:::::::,,jEEKKWWWWKKKKKKWWWWWWWWW##KWW#####KWWWWWWKKKKEi..D....................................:jEEE.........;KKWWWWWWWW#WW##W##########WWKKDLGKEKDDDDDDDDDDDDDDDDDDDDDDDDDED *
*jji. i:.::::::::,,,EEEKKWWWWKKKKKWWWWWWWWWW##WWW#####WWWWWWWKKKKKi..D......................................:::::......,KKKWWWWWWWWW#####W########WWWKKKGGKKEGGGGGGGGDDDDDDDDDDDDDDDDDDE *
*jji. i..:::::::::,,tEEKKWWWWKKKKKWWWWWWWWWWW##WW######WWWWWWWKKKKKi..D......................................::::......:EKKKWWWWWWWWWWW##WW########W#WKKWGGKKGGGGGGGGGGGGGGGDDDDDDDDDDDDD *
*jji. .:::::::::::,,,EEEKKWWWWKKKKKWWWWWWWWWWW##WW#####WWWWWWWWKKKKKi..D....................................:::::::::..tELii;KWWWWWWWWWW##WW######WWWWWWKWGGGKGGGGGGGGGGGGGGGGGGGGGGGGGGDG *
*jjt. :.::::::::,,,,fEEKKWWWWKKKKKKWWWWWWWWWW###WW####WWWWWWW#WKKKKKi..D....................................:::::::.:.,;;;;;;,KKWWWWWWWWW#WW########WWWKKWGGGKGGGGGGGGGGGGGGGGGGGGGGGGGGGG *
*jji. ;.::::::::,,,,;EEEKWWWWWKKKKKWWWWWWWWWWWW##WW###WKWWWWWK#WKKKKKi..G......................................:::::::,;;;;:...KKKWWWWWWWWWKWW#######WWWWKKGLGKDGGGGGGLLGGGGGGGGGGGGGGGGGGG *
*jjt. f.:::::::::,,,,fEEKKWWWWWKKKKKWWWWWWWWWWW###WW##WKKWWWWWW#WKKKKK;.jt........i.............................:::::::;j;;....:E.KKKWWWWWWWKWW#####W#WWWWKKLLGWEEGGGGGLGGGGGGGGGGGGGGGGGGGG *
*jjt. ...:::::::,,,,,;DEEKWWWWWKKKKKWWWWWWWWWWWW####WWWKKKWWWWWWWWKKKKK;.E;.........t.............................:::::ii;;.....D...KKWWWWWWWKWW#####WWEWWWKKGGGEKKGGGGGLGGGGGGGGGGGGGLGGGGGG *
*fji. ;.:::::::,,,,,;LEEKKWWWWWKKKKKWWWWWWWWWWWW####KWKKKKWWWWWWWWKKKKKi.D;..........j.............................:::tt;,.....:.....KKWWWWWWKWWWW##WWWGWWWKKGGGGKEGGGGGGGGGGGGGGGGGGGLLGGGGL *
*fji. t::::::::,,,,,,;EEEKWWWWWKKKKKKKWWWWWWWWWWW##WKWKKKKKWWWWWWWWKKKKKi:D;............j...........................::LL;,.............KKWWWWWKWWWWWWWWWGWWWKKGGGGKGGGGGGGGGGGGGGGGGGGGLLGGGGL *
*fjt: .:::::::,,,,,,,DEEKWWWWWWKKKKKKKWWWWWWWWWWWWKKWKKKKKKWWWWK#WWKKKKWitE;........... ............................:G;;:...............KKKWWKKWWWWWWWWWGWWWKKGGGGWGGGGGGGGGGGGGGGGGGGGGGGGGGL *
*fjji;:. .f:::::::,,,,,,,;EEEKWWWWWWKKKKKKWWWWWWWWWWWKKKKKKKKKKKWWKWWWWWKKKKWGKD;........................................L;;..................DKKWKKWWWWWWWWWGWWWKKDGGGKDGGGGGGGGGGGGGGGGGGGGGGGGGG *
*fjjtii;,:. :::::::,,,,,,,;EEEKWWWWWWKKKKKKWWWWWWWWWWKKKKKKKKKKKKWWWWWW#WWKKKKWiEj;......................................:i,;....,...............;KKEKWWWWWWWWWGKWWKKDDGGDEGGGDGGGGGDGGGGGGGGGGGGGGGG *
*fjtiiiii;;:. j::::::,,,,,,,;;EEEKWWWWW#KKKKKWWWWWWWWWKKKKKKWKKKKKKKWWWWWWWWWKKKKWtEL;:....................................;;;:...,;j................:KEEWWWWWWWWWDDWWKKDDDDDKDDDDDDDDDDDDDDDGGGGGGGGGGG *
*fjti;;iiii;;,:::::::,,,,,,,,;EEEKWWWWWWWKKKKWWWWWWWWKKKKKKKWKKKKKKKWWWWWWW#W#KKKKWEEii;...................................f;:....,;L...................EEKWWWWWWWWDDWWKKDDDDDKEDDDDDDDDDDDDDDDDDDDDDGGGG *
*fjt,,,;;;;ii;f::::::,,,,,,,;;EEKWWWWWWWKKKKKWWWKWWKKKKKKKKKKKKKKKKKWWWWWWW#W#KKKKWKEij;:...............................:G;,.....,;f....................:tKKWWWWWWWDDWWKKDDDDDKKDDDDDDDDDDDDDDDDDDDDDDDDD *
*jjt. ..:,;;;;,::::,,,,,,,,;;GEEWWWWWWWWKKKKWKKWKKKKKKKKKKKKKKKKKKKKWWWWWWW#W#KKKKWEDi;j;............................,Li;L;;;..,;;f........................KKKKWWWKDDWWKKDDDGDKKGGGGGGGGDGDDDDDDDDDDDDDDD *
*fjt. .:,,,:::::,,,,,,,;;;EEKWWWWWWWKKKKKKWKKKKKKKKKKKKKKKKKKKKKWKKKWKW##W#KKKKWEti;;G;........................tEEEL;;;;;;;;;;L..........................DKKKKKEDDWWKEDGftiLE;;;;itjLGGGGGGDDDDDDDDDDD *
*fjt. .j::::,,,,,,,;;;DEEWWWWWWWWKKKKWKKKKKKKKKKKKKKKKKKKKKKKWKKWWWK##W#KKKKKEii;;;L;...................iDEEEEEEKKi;j;;;;jD.....:......................,KKKKDGGEKKE:::::;E::::::::::,tLGGDDDDDDDDDD *
*fjt. .;:::,,,,,,,;;;;EEKWWWWWWWWKWKKKKKKKKKKKKKKKWKKKKKKKKKKWKKWWWW#WW#KKKKKKii;;;;f;.............:tDEEEEEKKKKKKKKEti;;;L...............................EEKf;:iKKE::::::E::::::::::::::ifDDDDDDDDD *
*fjt: :::,,,,,,,,;;;DEEWWWWWWWWWEKKKKKKKKKKKKKKKKKKKKKKKKKKWWKKWWWW####KKKKKEiii;;;;f,.........iDEEEEKKKKKKKKKKKKKKKf;iG......i..........................fK::::KKE::::::E::::::::::::::::,tGGDDDDD *
*fjt: t:::,,,,,,;;;;iDEKWWWWWWKEKKKKKKKKKKKKKKKKKKKKKKKKKKKKWWKKWWWW####WKKKKLiii;;;;;L,....,Li;EDEEEEKKKKKKKKKKKKKKKKiG......;:...........................:i:::KKE:::::,E,::::::::::::::::::iGDDDD *
*jjt. f::,,,,,,,;;;;GEEWWWWKEEKEKKKKKKKKKKKKKKKKWKKKKKKKKKKKWWKWWWWW###WWKKKKiii;;;;;;;G,;L;;iiEEEEEEEKKKKKKKKKKKKKWWKE......;t.........:....................j::KEE:,::,,D,,::::,,,,,,:::::::::tDDD *
*fjt:. ,::,,,,,,,;;;;EEWWKEEEEEEKKKKKKKKKKKKKKKKWKKKKKKKKKKKWWKKWWWWW#W#KKKKKKiiiiii;;;;;i;;iiiEEKEEKKWKKKKKKKWKKKKKWWWGi;...;t......,;;;;,....................:,EEE,,,,,,D,,,,,,,,,,,,,,,,::,::::tG *
*fjt:. ,::,,,,,,,;;;;DEKEEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKKKKKWWWWWWWWW#W#KKKKKKiiiii;;i;;;;;iiiKKKEKKKKWWKWWWWWWKKKKWWWWW;;;:;L.....;;;;;;;;;....................,KEE,,,,,,E,,,,,,,,,,,,,,,,,,,,,,,,; *
*fjt:. f:,,,,,,,;;;;jEDEEEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKKKKKWWWWWWWW#W##KKKKKKiiiiiiii;i;;iiiEKKKKKKKKWKWWWWWWWWKKKWWWWWKi;;i.....,jEEfGi;;;;;...................EED,,,,,,E,,,,,,,,,,,,,,,,,,,,,,,,, *
*fjt:. .f::,,,,,,;;jEEDEEEEEEEEEEKKKKKKKKKKKKKKKWKKKKKKKKKKKKKWWWKWWWWW###KKKKKLiiiiiiiiiiiiiiEEKKKKKKKKWWWWWWWWWWWWKWWWWWWGi;i;,..;jDDDKEGi;;;;;;:................EED,,,,,,D,,,,,,,,,,,,,,,,,,,,,,,,, *
*fjt:. .. ;::,,,,,;;EDDEEEEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKWWKKW#WW####KWKKKiiiiiiiiiiiiijKKKKKKKKKKWWWWWWWWWWWWWWWWWWWWWt;i;;;;i;DDDDDDGi;;;;;;;;:.............EDf;,,,;,G;;;;;;;;;;;;;;;,,,,,,,,,, *
*fjt:......:,,,,,,;LDDDEEEEEEEEEEEKKKKKKKKKKKKKKKKWKKKKKKKKKKKKKWWWWKWWWW####KKKKKiiiiiiiiiiijKEKKWKKKKKKKWWWWWWWWWWWWWWWWWWWWWWiLiii;i;DEEEEDDE;i;;;;;;;;;:..........EDi,;;;;;L;;;;;;;;;;;;;;;;;;,,,,,,, *
*fjt:......:,,,,,;EDDDEEKEEEEEEEEEKKKKKKKKKKKKKKKWKKKKKKKKKKKKKKWWWWKKWWW##W#KWKKWEiiiiiijGKKKKKWWKKKKKKKKWWWWWWWWWWWWWWWWWWWWWWKi;iiiiDDEEEEEEDEi;;;;;;;;;;;;;,:.....ED;;;;;;;j;;;;;;;;;;;;;;;;;;;;;;;,, *
*fjt:.....t,,,,,;DDDDEEEKEEEEEEEEKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKWWWKKKWWWW##WKWKKWKiiiKKKKKKKKKWWKKKKKKKKWWWWWWWWWWWWWWW#WWWWWWWWWiiiiifLEEEEEEEEDi;i;;;;;;;;;;;;.....DD;;;;;;;i;;;;;;;;;;;;;;;;;;;;;;;;; *
*fjt:.....G,,,,,GDDDEEEEEEEEEEEEKKKKKKKKKKKKKKKKWKKKKKKKKKKKKKKKWWWKKKWWW###WKWKKWKitKKKKKKKKKWKKKKKKKKKKWWWWWWWWWWWWWW###WWWWWWWWEiiiiiiiEEEEEEEEDGiiii;;;;;;;;;.....GD;;;;;;;i;;;;;;;;;;;;;;;;;;;;;;;;; *
*fjt:.....L,,,,;GDDDEEEEEEEEEEKEKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKWWWWWDGWWW###KKWWKWKKKKKKKKKKKKKKKKKKKKKKKWWWWWWWWWWWWW####WWWWWWWWWiiiiiiiiEEEEEEEEEEDi;i;;;;;;;;.....Lj;;;;;;i;iiiiii;;;;;;ii;;;;;;;;;;; *
***********************************************************************************************************************************************************************************************************/
// Classes
static class Edge implements Comparable<Edge>{
int l,r;
Edge(){}
Edge(int l,int r){
this.l=l;
this.r=r;
}
@Override
public int compareTo(Edge e) {
return (l-e.l)!=0?l-e.l:r-e.r;
}
}
static class Segment implements Comparable<Segment> {
long l, r, initialIndex;
Segment () {}
Segment (long l_, long r_, long d_) {
this.l = l_;
this.r = r_;
this.initialIndex = d_;
}
@Override
public int compareTo(Segment o) {
return (int)((l - o.l) !=0 ? l-o.l : initialIndex - o.initialIndex);
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st==null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String s="";
try {
s=br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return s;
}
}
// Functions
static long gcd(long a, long b) {
return b==0?a:gcd(b,a%b);
}
static int gcd(int a, int b) {
return b==0?a:gcd(b,a%b);
}
static void reverse(long[] A,int l,int r) {
int i=l,j=r-1;
while(i<j) {
long t=A[i];
A[i]=A[j];
A[j]=t;
i++;j--;
}
}
static void reverse(int[] A,int l,int r) {
int i=l,j=r-1;
while(i<j) {
int t=A[i];
A[i]=A[j];
A[j]=t;
i++;j--;
}
}
//Input Arrays
static void input(long a[], int n) {
for(int i=0;i<n;i++) {
a[i]=inp.nextLong();
}
}
static void input(int a[], int n) {
for(int i=0;i<n;i++) {
a[i]=inp.nextInt();
}
}
static void input(String s[],int n) {
for(int i=0;i<n;i++) {
s[i]=inp.next();
}
}
static void input(int a[][], int n, int m) {
for(int i=0;i<n;i++) {
for(int j=0;j<m;j++) {
a[i][j]=inp.nextInt();
}
}
}
static void input(long a[][], int n, int m) {
for(int i=0;i<n;i++) {
for(int j=0;j<m;j++) {
a[i][j]=inp.nextLong();
}
}
}
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class r653p5 {
private static PrintWriter pw = new PrintWriter(System.out);
private static InputReader sc = new InputReader();
private static final int a = 0, b = 1, ab = 2;
static class InputReader{
private static BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
private static StringTokenizer tk;
private void next()throws IOException{
if(tk == null || !tk.hasMoreTokens())
tk = new StringTokenizer(r.readLine());
}
private int nextInt()throws IOException{
next();
return Integer.parseInt(tk.nextToken());
}
}
public static void main(String args[])throws IOException{
solve();
pw.flush();
pw.close();
}
private static void solve()throws IOException{
int n = sc.nextInt(), k = sc.nextInt();
ArrayList<ArrayList<Integer>> list = new ArrayList<>();
int size[] = new int[3];
for(int i=0; i<3; i++)
list.add(new ArrayList<>());
for(int i=0; i<n; i++){
int t = sc.nextInt(), ch1 = sc.nextInt(), ch2 = sc.nextInt();
int q = ch1 * 2 + ch2 - 1;
if(q < 0) continue;
list.get(q).add(t);
size[q]++;
}
for(int i=0; i<3; i++)
Collections.sort(list.get(i));
long sum[][] = new long[3][];
for(int i=0; i<3; i++){
sum[i] = new long[size[i]+1];
for(int j=0; j<size[i]; j++)
sum[i][j+1] = sum[i][j] + list.get(i).get(j);
}
long ans = Long.MAX_VALUE;
for(int q=0; q<=k; q++){
if(q <= size[ab] && k-q <= Math.min(size[a], size[b])){
long temp = sum[ab][q] + sum[a][k-q] + sum[b][k-q];
ans = Math.min(temp, ans);
}
}
ans = (ans == Long.MAX_VALUE) ? -1 : ans;
pw.println(+ans);
}
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
n,k=map(int,input().split())
l,l1,l2=[],[],[]
for i in range(n):
t,a,b=map(int,input().split())
if a==1 and b==1:
l.append(t)
if a==1 and b==0:
l1.append(t)
if a==0 and b==1:
l2.append(t)
l1.sort()
l2.sort()
for i in range(min(len(l1),len(l2))):
l.append(l1[i]+l2[i])
l.sort()
if len(l)<k:
print(-1)
else:
print(sum(l[:k]))
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
// package com.company;
import java.util.*;
import java.lang.*;
import java.io.*;
//****Use Integer Wrapper Class for Arrays.sort()****
public class DR5 {
static PrintWriter out=new PrintWriter(new OutputStreamWriter(System.out));
public static void main(String[] Args)throws Exception{
FastReader scan=new FastReader(System.in);
int t=1;
// t=scan.nextInt();
while(t-->0){
int n=scan.nextInt();
int k=scan.nextInt();
book[] arr=new book[n];
int al=0;
int bl=0;
for(int i=0;i<n;i++){
int ti=scan.nextInt();
int a=scan.nextInt();
int b=scan.nextInt();
if(a==1){
al+=1;
}if(b==1){
bl+=1;
}
arr[i]=new book(ti,a,b);
}
if(al<k||bl<k){
out.println(-1);
}else{
Arrays.sort(arr);
ArrayList<Integer> oal=new ArrayList<>();
ArrayList<Integer> obl=new ArrayList<>();
ArrayList<Integer> abl=new ArrayList<>();
for(int i=0;i<n;i++){
book cur=arr[i];
if(cur.a==1&&cur.b==1){
abl.add(i);
}else{
if(cur.a==1){
oal.add(i);
}if(cur.b==1){
obl.add(i);
}
}
}
int ap=0;
int bp=0;
int cp=0;
int an=k-Math.min(k,oal.size());
int bn=k-Math.min(k,obl.size());
cp=Math.max(an,bn);
long ans=0;
for(int i=0;i<cp;i++){
ans+=arr[abl.get(i)].t;
k--;
}
while(k>0){
long at=arr[oal.get(ap)].t;
long bt=arr[obl.get(bp)].t;
long ct=Integer.MAX_VALUE;
if(cp<abl.size()){
ct=arr[abl.get(cp)].t;
}
if(at+bt<=ct){
ans+=at+bt;
ap++;
bp++;
}else{
ans+=ct;
cp++;
}
k--;
}
out.println(ans);
}
}
out.flush();
out.close();
}
static class book implements Comparable<book>{
int t;
int a;
int b;
book(int t,int a,int b){
this.t=t;
this.a=a;
this.b=b;
}
@Override
public int compareTo(book o) {
return this.t-o.t;
}
}
static class FastReader {
byte[] buf = new byte[2048];
int index, total;
InputStream in;
FastReader(InputStream is) {
in = is;
}
int scan() throws IOException {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0) {
return -1;
}
}
return buf[index++];
}
String next() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan()) ;
StringBuilder sb = new StringBuilder();
for (; c > 32; c = scan()) {
sb.append((char) c);
}
return sb.toString();
}
int nextInt() throws IOException {
int c, val = 0;
for (c = scan(); c <= 32; c = scan()) ;
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
long nextLong() throws IOException {
int c;
long val = 0;
for (c = scan(); c <= 32; c = scan()) ;
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
}
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
import sys
def swap(x, y, a):
temp = a[x]
a[x] = a[y]
a[y] = temp
return a
def solve():
return None
def main():
q = []
for line in sys.stdin.readlines():
q.append(line)
for i in range(len(q)):
q[i] = q[i].rstrip().split(' ')
q[i] = [int(x) for x in q[i]]
k = q[0][1]
# combining the ones that we can combine
# get rid of the ones where both people dont like reading
both = [] # keeps track of books both people like
only_left = [] # only left person likes
only_right = [] # only right person likes
for i in range(1,len(q)):
if q[i][1] == 1 and q[i][2] == 1:
both.append(q[i][0])
elif q[i][1] == 1 and q[i][2] == 0:
only_left.append(q[i][0])
elif q[i][1] == 0 and q[i][2] == 1:
only_right.append(q[i][0])
# print(both)
# print(only_left)
# print(only_right)
only_left.sort()
only_right.sort()
for i in range(min(len(only_left), len(only_right))):
both.append(only_left[i] + only_right[i])
both.sort()
if len(both) < k:
print(-1)
else:
# print(both)
print(sum(both[:k]))
# print(k, q)
if __name__ == '__main__':
main()
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
line = input()
n, m, k = [int(i) for i in line.split(' ')]
books, allL, aliceL, bobL, other =list(range(1, n + 1)), [], [], [], []
ts = [[] for _ in range(n + 1)]
for i in range(n):
line = input()
t, a, b = [int(j) for j in line.split(' ')]
ts[i + 1] = [t, a, b]
if a == 1 and b == 1:
allL.append(i + 1)
elif a == 1:
aliceL.append(i + 1)
elif b == 1:
bobL.append(i + 1)
else:
other.append(i + 1)
if len(allL) + min(len(aliceL), len(bobL)) < k or (len(allL) < k and 2 * (k - len(allL)) > m - len(allL)) :
print(-1)
exit()
books.sort(key=lambda x: (ts[x][0], -ts[x][2]*ts[x][1]))
allL.sort(key=lambda x: (ts[x][0], -ts[x][2]*ts[x][1]))
aliceL.sort(key=lambda x: (ts[x][0], -ts[x][2]*ts[x][1]))
bobL.sort(key=lambda x: (ts[x][0], -ts[x][2]*ts[x][1]))
other.sort(key=lambda x: (ts[x][0], -ts[x][2]*ts[x][1]))
x = max(2 * k - m, 0, k - min(len(aliceL), len(bobL)), m - len(aliceL) - len(bobL) - len(other))
cura, curb, curo, cur = max(0, k - x), max(0, k - x), 0, sum(ts[i][0] for i in allL[:x])
cur += sum(ts[i][0] for i in aliceL[:cura]) + sum(ts[i][0] for i in bobL[:curb])
while cura + x + curb + curo < m:
an = ts[aliceL[cura]][0] if cura < len(aliceL) else 9999999999
bn = ts[bobL[curb]][0] if curb < len(bobL) else 9999999999
on = ts[other[curo]][0] if curo < len(other) else 9999999999
cur += min(an, bn, on)
if an <= bn and an <= on:
cura += 1
elif bn <= an and bn <= on:
curb += 1
else:
curo += 1
res, a, b, o = cur, cura, curb, curo
for i in range(x + 1, len(allL) + 1): #ι½εζ¬’ηιΏεΊ¦
cur += ts[allL[i - 1]][0]
if cura > 0:
cura -= 1
cur -= ts[aliceL[cura]][0]
if curb > 0:
curb -= 1
cur -= ts[bobL[curb]][0]
if curo > 0:
curo -= 1
cur -= ts[other[curo]][0]
while cura + i + curb + curo < m:
an = ts[aliceL[cura]][0] if cura < len(aliceL) else 9999999999
bn = ts[bobL[curb]][0] if curb < len(bobL) else 9999999999
on = ts[other[curo]][0] if curo < len(other) else 9999999999
cur += min(an, bn, on)
if an <= bn and an <= on:
cura += 1
elif bn <= an and bn <= on:
curb += 1
else:
curo += 1
if res > cur:
res,x, a, b, o = cur,i, cura, curb, curo
print(res)
for i in range(x):
print(allL[i], end=' ')
for i in range(a):
print(aliceL[i], end = ' ')
for i in range(b):
print(bobL[i], end = ' ')
for i in range(o):
print(other[i], end = ' ')
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
n,k = map(int,input().split())
a=[]
b=[]
c=[]
d=[]
for i in range(n):
x,y,z = map(int,input().split())
if y==0 and z==0:
d.append(x)
elif y==1 and z==1:
a.append(x)
elif y==0 and z==1:
c.append(x)
else:
b.append(x)
b.sort()
c.sort()
m=min(len(b),len(c))
for i in range(m):
a.append(b[i]+c[i])
a.sort()
if len(a)<k:
print("-1")
else:
print(sum(a[:k]))
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python2
|
#!/usr/bin/env python
# -*- encoding: utf-8 -*-
'''
@File : D.py.py
@Contact : wangweixu@zuoyebang.com,liuyang05@zuoyebang.com,wangqian03@zuoyebang.com
@License : (C)Copyright 2020-2022; All rights reserved.
@Modify Time @Author @Version @Desciption
------------ ------- -------- -----------
2020/10/23 4:30 δΈε δΊ§εθ―΄ηι½ι 1.0 None
'''
import sys
a=[]
b=[]
c=[]
n,m=sys.stdin.readline().strip().split(" ")
n=int(n)
m=int(m)
for i in range(n):
x,y,z=sys.stdin.readline().strip().split(" ")
if y=="1" and z=="1":
a.append(int(x))
if y=="1" and z=="0":
b.append(int(x))
if y=="0" and z=="1":
c.append(int(x))
a.sort()
b.sort()
c.sort()
i,j,k=0,0,0
res=0
while m>0:
if (i>=len(a)) and ((j>=len(b)) or (k>=len(c))):
print -1
exit(0)
if i>=len(a):
t1=-1
else:
t1=a[i]
if j>=len(b):
t2=-1
else:
t2=b[j]
if k>=len(c):
t3= -1
else:
t3=c[k]
if (t1==-1):
ss=t2+t3
res+=ss
j+=1
k+=1
m-=1
else:
if (t2 == -1) or (t3==-1):
res+=t1
i+=1
m-=1
else:
ss = t2+t3
if t1>ss:
res+=ss
m-=1
j+=1
k+=1
else:
res+=t1
m-=1
i+=1
print res
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios {
fast_ios() {
cin.tie(0);
ios::sync_with_stdio(false);
cout << fixed << setprecision(20);
};
} fast_ios_;
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
for (auto &v : vec) is >> v;
return is;
}
template <typename T>
bool chmax(T &m, const T q) {
if (m < q) {
m = q;
return true;
} else
return false;
}
template <typename T>
bool chmin(T &m, const T q) {
if (q < m) {
m = q;
return true;
} else
return false;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <class c>
struct rge {
c b, e;
};
template <class c>
rge<c> range(c i, c j) {
return rge<c>{i, j};
}
template <class c>
auto dud(c *x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct deb {
~deb() { cerr << endl; }
template <class c>
typename enable_if<sizeof dud<c>(0) != 1, deb &>::type operator<<(c i) {
cerr << boolalpha << i;
return *this;
}
template <class c>
typename enable_if<sizeof dud<c>(0) == 1, deb &>::type operator<<(c i) {
return *this << range(begin(i), end(i));
}
template <class c, class b>
deb &operator<<(pair<b, c> d) {
return *this << "(" << d.first << ", " << d.second << ")";
}
template <class c>
deb &operator<<(rge<c> d) {
*this << "[";
for (auto it = d.b; it != d.e; ++it) *this << ", " + 2 * (it == d.b) << *it;
return *this << "]";
}
};
void err(istream_iterator<string> it) { return; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
cerr << *it << " = " << a << endl;
err(++it, args...);
}
void solve() {
lint n, k;
cin >> n >> k;
vector<lint> time(n);
vector<lint> alice(n);
vector<lint> bob(n);
priority_queue<int, vector<int>, greater<int>> abra;
priority_queue<int, vector<int>, greater<int>> cadabra;
priority_queue<int, vector<int>, greater<int>> mixed;
for (int i = (0), i_end_ = (n); i < i_end_; i++) {
lint t, a, b;
cin >> t >> a >> b;
time[i] = t;
alice[i] = a;
bob[i] = b;
if (a == 1 && b == 1) {
mixed.push(t);
} else if (a == 1) {
abra.push(t);
} else if (b == 1) {
cadabra.push(t);
}
}
lint ret = 0;
lint cc = 2 * k;
lint s1 = accumulate(begin(alice), end(alice), 0LL);
lint s2 = accumulate(begin(bob), end(bob), 0LL);
if (s1 <= k - 1 || s2 <= k - 1) {
cout << -1 << '\n';
return;
}
while (cc) {
bool ok = 0;
int f = 0;
lint t = INT_MAX;
if (mixed.size() > 0) {
t = mixed.top();
ok = 1;
}
lint t1 = INT_MAX;
lint t2 = INT_MAX;
if (abra.size()) {
t1 = abra.top();
++f;
}
if (cadabra.size()) {
t2 = cadabra.top();
++f;
}
if (!ok && f <= 1) break;
if (t <= t1 + t2) {
ret += t;
mixed.pop();
} else {
ret += t1 + t2;
abra.pop();
cadabra.pop();
}
cc -= 2;
}
cout << ret << '\n';
}
int main() {
int T = 1;
for (int tt = 1; tt <= T; ++tt) {
solve();
}
return 0;
}
lint pow_mod(lint base, lint expo) {
lint ret = 1;
for (; expo;) {
if (expo & 1) ret = (ret * base) % 1000000007;
expo = (expo >> 1);
base = (base * base) % 1000000007;
}
return ret;
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
# cook your dish here
# input = open('file.txt').readline
n , k = list(map(int , input().split()))
ali = []
bob = []
both = []
ac , bc = 0 , 0
for __ in range(n):
t , ai , bi = list(map(int , input().split()))
if ai == bi and ai == 1:
both.append(t)
ac += 1
bc += 1
elif ai == 1:
ali.append(t)
ac += 1
elif bi == 1:
bob.append(t)
bc += 1
if ac < k or bc < k:
print(-1)
else:
ans = 0
i = 0
both.sort()
ali.sort()
bob.sort()
bbi = 0
ai , bi = 0 , 0
bbl = len(both)
al = len(ali)
bl = len(bob)
while i < k:
if bbi < bbl and ai < al and bi < bl:
age = both[bbi]
pic = ali[ai] + bob[bi]
if age < pic:
ans += age
# both.pop(0)
bbi += 1
else:
ans += pic
# ali.pop(0)
# bob.pop(0)
ai += 1
bi += 1
elif bbi < bbl:
ans += both[bbi]
bbi += 1
else:
ans += ( ali[ai] + bob[bi] )
ai += 1
bi += 1
i += 1
print(ans)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
n, k = map(int, input().split())
al=[];bob=[];both=[]
for i in range(n):
a,b,c= map(int, input().split())
if(b==1 and c==1):
both.append(a)
elif(b==1):
al.append(a)
elif(c==1):
bob.append(a)
tp=len(both)
la=len(al)
ob=len(bob)
a=la+tp
b=ob+tp
if(a<k or b<k):
print("-1")
else:
both.sort()
al.sort()
bob.sort()
ans=0
c=0;d=0
for i in range(k):
z=20001
m=20001
if(c<tp):
z=both[c]
if(d<la and d<ob):
m=(al[d]+bob[d])
if(z<m):
ans+=z
c+=1
else:
ans+=m
d+=1
print(ans)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
inline long long gcd(long long a, long long b) {
while (b != 0) {
long long c = a % b;
a = b;
b = c;
}
return a < 0 ? -a : a;
}
inline long long lowbit(long long x) { return x & (-x); }
const double PI = 3.14159265358979323846;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long mod = 998244353;
inline long long rd() {
long long x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
return x * f;
}
const double eps = 1e-6;
const int M = 1e6 + 10;
const int N = 1e6 + 10;
struct MM {
int id, val;
bool friend operator<(const MM& m1, const MM& m2) { return m1.val < m2.val; }
};
MM a[N], b[N], c[N], d[N];
int x1 = 0, x2 = 0, x3 = 0, x4 = 0;
int n, m, k;
vector<int> ans;
MM pp[N];
long long f(int x) {
ans.clear();
long long sum = 0;
if (x > x3 || x + x1 < k || x + x2 < k || 2 * k - x > m) return INF;
for (int i = 1; i <= x; i++) {
sum += c[i].val;
ans.push_back(c[i].id);
}
for (int i = 1; i <= k - x; i++) {
sum += a[i].val;
ans.push_back(a[i].id);
}
for (int i = 1; i <= k - x; i++) {
sum += b[i].val;
ans.push_back(b[i].id);
}
int cnt = 0;
for (int i = max(k - x + 1, 1); i <= x1; i++) {
pp[++cnt] = a[i];
}
for (int i = max(k - x + 1, 1); i <= x2; i++) {
pp[++cnt] = b[i];
}
for (int i = x + 1; i <= x3; i++) {
pp[++cnt] = c[i];
}
for (int i = 1; i <= x4; i++) {
pp[++cnt] = d[i];
}
int res = m - (2 * k - x);
sort(pp + 1, pp + 1 + cnt);
for (int i = 1; i <= res; i++) {
ans.push_back(pp[i].id);
sum += pp[i].val;
}
return sum;
}
int main() {
n = rd(), m = rd(), k = rd();
for (int i = 1; i <= n; i++) {
int t = rd(), aa = rd(), bb = rd();
if (aa && bb)
c[++x3] = {i, t};
else if (aa)
a[++x1] = {i, t};
else if (bb)
b[++x2] = {i, t};
else
d[++x4] = {i, t};
}
sort(a + 1, a + 1 + x1);
sort(b + 1, b + 1 + x2);
sort(c + 1, c + 1 + x3);
sort(d + 1, d + 1 + x4);
if (x3 + x1 < k || x3 + x2 < k) {
cout << -1 << endl;
return 0;
}
int l = 0, r = x3;
int an = l;
while (l + 10 < r) {
int lmid = l + (r - l) / 3, rmid = r - (r - l) / 3;
if (f(lmid) < f(rmid)) {
an = lmid;
r = rmid - 1;
} else {
an = rmid;
l = lmid + 1;
}
}
for (int i = l; i <= r; i++) {
if (f(i) < f(an)) an = i;
}
long long aa = f(an);
if (aa == INF) {
cout << -1 << endl;
return 0;
}
cout << aa << endl;
for (auto& x : ans) {
printf("%d ", x);
}
cout << endl;
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
struct books {
int t, i;
};
bool cmp(books a, books b) {
if (a.t != b.t) return (a.t < b.t);
return (a.i < b.i);
}
int main() {
int n, m, o, x, y, i, j, t, s, k, l, p = 0, q, e[200001], f[200001];
books a[200001], b[200001], c[200001], d[200001];
long long ans;
cin >> n >> o >> m;
i = 0;
j = 0;
k = 0;
l = 0;
for (q = 1; q <= n; q++) {
cin >> t >> x >> y;
if (x == 1 && y == 1) {
c[k].t = t;
c[k++].i = q;
} else if (x == 1) {
a[i].t = t;
a[i++].i = q;
} else if (y == 1) {
b[j].t = t;
b[j++].i = q;
} else {
d[l].t = t;
d[l++].i = q;
}
}
sort(a, a + i, cmp);
sort(b, b + j, cmp);
sort(c, c + k, cmp);
sort(d, d + l, cmp);
x = 0;
y = 0;
t = 0;
s = 0;
ans = 0;
o -= m;
while (m > 0) {
if (x < i && y < j && t < k) {
if (a[x].t + b[y].t <= c[t].t && o > 0) {
ans += a[x].t + b[y].t;
e[p++] = a[x].i;
e[p++] = b[y].i;
x++;
y++;
o--;
} else {
ans += c[t].t;
e[p++] = c[t].i;
t++;
}
m--;
} else if ((x == i || y == j) && t < k) {
ans += c[t].t;
e[p++] = c[t].i;
t++;
m--;
} else if (t == k && x < i && y < j && o > 0) {
ans += a[x].t + b[y].t;
e[p++] = a[x].i;
e[p++] = b[y].i;
x++;
y++;
m--;
o--;
} else
break;
}
if (m == 0) {
q = 0;
while (o--) {
m = 10001;
if (x < i && a[x].t < m) m = a[x].t;
if (y < j && b[y].t < m) m = b[y].t;
if (t < k && c[t].t < m) m = c[t].t;
if (s < l && d[s].t < m) m = d[s].t;
if (t > 0 && x < i && y < j && c[t - 1].t + m > a[x].t + b[y].t) {
ans += a[x].t + b[y].t;
e[p++] = a[x].i;
e[p++] = b[y].i;
x++;
y++;
ans -= c[t - 1].t;
f[q++] = c[t - 1].i;
t--;
} else {
ans += m;
if (x < i && a[x].t == m) {
e[p++] = a[x].i;
x++;
} else if (y < j && b[y].t == m) {
e[p++] = b[y].i;
y++;
} else if (t < k && c[t].t == m) {
e[p++] = c[t].i;
t++;
} else if (s < l && d[s].t == m) {
e[p++] = d[s].i;
s++;
}
}
}
cout << ans << endl;
sort(e, e + p);
sort(f, f + q);
y = 0;
for (x = 0; x < p; x++) {
if (y < q && e[x] == f[y])
y++;
else
cout << e[x] << " ";
}
cout << endl;
} else
cout << -1 << endl;
return 0;
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
s=input()
s1=s.split()
n=int(s1[0])
k=int(s1[1])
import heapq
heap1=[]
heap2=[]
heap3=[]
heapq.heapify(heap1)
heapq.heapify(heap2)
heapq.heapify(heap3)
for i in range(n):
s=input()
s1=s.split()
if int(s1[1])==0 and int(s1[2])==1:
heapq.heappush(heap1,int(s1[0]))
if int(s1[1])==1 and int(s1[2])==0:
heapq.heappush(heap2,int(s1[0]))
if int(s1[1])==1 and int(s1[2])==1:
heapq.heappush(heap3,int(s1[0]))
sum1=0
while ((heap1 and heap2) or heap3) and k>0:
if heap1 and heap2:
c1=heapq.heappop(heap1)
c2=heapq.heappop(heap2)
else:
c1=1e10
c2=1e10
if heap3:
c3=heapq.heappop(heap3)
else:
c3=1e10
if c3<(c2+c1):
sum1=sum1+c3
if c1!=1e10:
heapq.heappush(heap1,c1)
if c2!=1e10:
heapq.heappush(heap2,c2)
else:
sum1=sum1+c2+c1
if c3!=1e10:
heapq.heappush(heap3,c3)
k=k-1
if k>0:
print(-1)
else:
print(sum1)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
def compute(n, k, arr):
add = 0
both = []
alice = []
bob = []
for i in range(n):
if arr[i][1] and arr[i][2]:
both.append(arr[i][0])
elif arr[i][1]:
alice.append(arr[i][0])
elif arr[i][2]:
bob.append(arr[i][0])
both.sort()
alice.sort()
bob.sort()
#print(alice, bob, both)
if (len(alice)+len(both) < k) or (len(both)+len(bob) < k):
return -1
else:
i, j = 0, 0
n, nn, nnn = len(alice), len(bob), len(both)
for _ in range(k):
if not alice or not bob or j >= n or j > nn or (i < nnn and both[i] <= alice[j] + bob[j]):
add += both[i]
i += 1
else:
add += (alice[j] + bob[j])
j += 1
return add
if __name__ == "__main__":
n, k = map(int, input().split())
arr = []
for _ in range(n):
t, a, b = map(int, input().split())
arr.append([t, a, b])
print(compute(n, k, arr))
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
/*input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
*/
import java.math.*;
import java.io.*;
import java.util.*;
public class Main
{
static class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
static Reader sc=new Reader();
static BufferedWriter bw=new BufferedWriter(new OutputStreamWriter(System.out));
public static void main(String args[])throws IOException
{
/*
* For integer input: int n=inputInt();
* For long input: long n=inputLong();
* For double input: double n=inputDouble();
* For String input: String s=inputString();
* Logic goes here
* For printing without space: print(a+""); where a is a variable of any datatype
* For printing with space: printSp(a+""); where a is a variable of any datatype
* For printing with new line: println(a+""); where a is a variable of any datatype
Scanner in = new Scanner(System.in);
//all four
int[] dr = { 1, 0, -1, 0 };
int[] dc = { 0, 1, 0, -1 };
*/
int n=inputInt();
int k=inputInt();
PriorityQueue<Integer> pcom=new PriorityQueue<>();
PriorityQueue<Integer> p1=new PriorityQueue<>();
PriorityQueue<Integer> p2=new PriorityQueue<>();
int cnt1=k,cnt2=k;
for(int i=0;i<n;i++)
{
int t=inputInt(),o=inputInt(),z=inputInt();
if(o==1 && z==1)
{
pcom.add(t);
}
else if(o==0 && z==1)
{
p2.add(t);
}
else if(o==1 && z==0)
{
p1.add(t);
}
}
while(!p1.isEmpty() && !p2.isEmpty())
{
pcom.add(p1.poll()+p2.poll());
}
if(pcom.size()<k)
println(-1+"");
else
{
int ans=0;
while(k-->0)
{
ans+=pcom.poll();
}
println(ans+"");
}
bw.flush();
bw.close();
}
static long nCr(long n, long r)
{
long ans=1;
if(r>n-r)
{
r=n-r;
}
for(long i=1;i<=r;i++)
{
ans*=(n-i+1);
ans/=i;
}
return ans;
}
public static int modulo(int x,int N)
{
return (x % N + N) %N;
}
public static long lcm(long a,long b)
{
return a / gcd(a, b) * b;
}
public static long gcd(long a,long b)
{
if(b==0)
return a;
return gcd(b,a%b);
}
public static int inputInt()throws IOException
{
return sc.nextInt();
}
public static long inputLong()throws IOException
{
return sc.nextLong();
}
public static double inputDouble()throws IOException
{
return sc.nextDouble();
}
public static String inputString()throws IOException
{
return sc.readLine();
}
public static void print(String a)throws IOException
{
bw.write(a);
}
public static void printSp(String a)throws IOException
{
bw.write(a+" ");
}
public static void println(String a)throws IOException
{
bw.write(a+"\n");
}
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
n,k = map(int,input().split())
aa=[]
bb=[]
l=[]
for _ in range(n):
t,a,b=map(int,input().split())
sum1=0
sum2=0
if a&b:
l.append(t)
elif a==1:
aa.append(t)
elif b==1:
bb.append(t)
bb.sort()
aa.sort()
nn=min(len(aa),len(bb))
for i in range(nn):
l.append(bb[i]+aa[i])
if k> len(l):
print(-1)
else:
l.sort()
sum1=0
for i in range(k):
sum1+=l[i]
print(sum1)
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
java
|
//Problem: Reading Books (easy version) (https://codeforces.com/problemset/problem/1374/E1)
//Status: Not accepted
import java.util.*;
import java.io.*;
public class _1374E1 {
private static BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
private static StringTokenizer st;
public static void main(String[] args) throws IOException {
st = new StringTokenizer(infile.readLine());
int numBooks = Integer.parseInt(st.nextToken());
int minLiked = Integer.parseInt(st.nextToken());
LinkedList<Book> both = new LinkedList<Book>();
LinkedList<Book> alice = new LinkedList<Book>();
LinkedList<Book> bob = new LinkedList<Book>();
int numAlice = 0, numBob = 0;
for(int i = 0; i < numBooks; i++) {
st = new StringTokenizer(infile.readLine());
Book temp = new Book(Integer.parseInt(st.nextToken()), Integer.parseInt(st.nextToken()), Integer.parseInt(st.nextToken()));
if(temp.alice && temp.bob)
both.add(temp);
else if(temp.alice)
alice.add(temp);
else if(temp.bob)
bob.add(temp);
if(temp.alice)
numAlice++;
if(temp.bob)
numBob++;
}
Collections.sort(both);
Collections.sort(alice);
Collections.sort(bob);
if(numAlice < minLiked || numBob < minLiked)
System.out.println("-1");
else {
numAlice = 0;
numBob = 0;
int minTime = 0;
while(numAlice < minLiked || numBob < minLiked) {
if(!both.isEmpty() && !alice.isEmpty() && !bob.isEmpty()) {
numAlice++;
numBob++;
if(both.get(0).time < alice.get(0).time + bob.get(0).time)
minTime += both.removeFirst().time;
else
minTime += alice.removeFirst().time + bob.removeFirst().time;
}
else if(!both.isEmpty() && (alice.isEmpty() || bob.isEmpty())) {
numAlice++;
numBob++;
minTime += both.removeFirst().time;
}
else if(both.isEmpty()) {
if(numAlice < minLiked) {
numAlice++;
minTime += alice.removeFirst().time;
}
if(numBob < minLiked) {
numBob++;
minTime += bob.removeFirst().time;
}
}
}
System.out.println(minTime);
}
}
}
class Book implements Comparable<Book> {
public boolean alice, bob;
public int time;
public Book(int time, int alice, int bob) {
this.time = time;
this.alice = alice == 1;
this.bob = bob == 1;
}
public int compareTo(Book other) {
return this.time - other.time;
}
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 7;
long long poww(long long a, long long b, long long mod) {
long long ans = 1;
while (b) {
if (b & 1) ans = (ans * a) % mod;
a = (a * a) % mod;
b >>= 1;
}
return ans % mod;
}
int main() {
int n, k;
cin >> n >> k;
vector<int> a;
vector<int> b;
vector<int> c;
vector<int> d;
for (int i = 0; i < n; i++) {
int z, x, y;
cin >> z >> x >> y;
if (x == 1 && y == 1) {
a.push_back(z);
} else if (x == 1) {
b.push_back(z);
} else if (y == 1) {
c.push_back(z);
}
}
if (a.size() + min(b.size(), c.size()) < k) {
cout << -1 << endl;
return 0;
}
sort(b.begin(), b.end());
sort(c.begin(), c.end());
for (int i = 0; i < b.size() && i < c.size(); i++) {
int z = b[i] + c[i];
a.push_back(z);
}
sort(a.begin(), a.end());
int sum = 0;
for (int i = 0; i < k; i++) {
sum += a[i];
}
cout << sum << endl;
}
|
1374_E1. Reading Books (easy version)
|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: t_i β the amount of time Alice and Bob need to spend to read it, a_i (equals 1 if Alice likes the i-th book and 0 if not), and b_i (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
* Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
* the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of t_i over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1 β€ k β€ n β€ 2 β
10^5).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers t_i, a_i and b_i (1 β€ t_i β€ 10^4, 0 β€ a_i, b_i β€ 1), where:
* t_i β the amount of time required for reading the i-th book;
* a_i equals 1 if Alice likes the i-th book and 0 otherwise;
* b_i equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T β the minimum total reading time of the suitable set of books.
Examples
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1
|
{
"input": [
"8 4\n7 1 1\n2 1 1\n4 0 1\n8 1 1\n1 0 1\n1 1 1\n1 0 1\n3 0 0\n",
"5 2\n6 0 0\n9 0 0\n1 0 1\n2 1 1\n5 1 0\n",
"5 3\n3 0 0\n2 1 0\n3 1 0\n5 0 1\n3 0 1\n"
],
"output": [
"18\n",
"8\n",
"-1\n"
]
}
|
{
"input": [
"2 1\n7 1 1\n2 1 1\n",
"5 1\n2 1 0\n2 0 1\n1 0 1\n1 1 0\n1 0 1\n",
"6 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 1\n3 0 1\n3 1 0\n3 0 0\n",
"6 3\n7 1 1\n8 0 0\n9 1 1\n6 1 0\n10 1 1\n5 0 0\n",
"8 4 3\n1 1 1\n3 1 1\n12 1 1\n12 1 1\n4 0 0\n4 0 0\n5 1 0\n5 0 1\n",
"6 3 1\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"3 3 1\n27 0 0\n28 0 0\n11 0 0\n",
"1 1 1\n3 0 1\n",
"8 5 1\n43 0 1\n5 0 1\n23 1 1\n55 0 1\n19 1 1\n73 1 1\n16 1 1\n42 1 1\n",
"6 3 2\n6 0 0\n11 1 0\n9 0 1\n21 1 1\n10 1 0\n8 0 1\n",
"9 2 2\n74 0 0\n78 1 0\n21 1 0\n47 1 0\n20 0 0\n22 0 1\n52 0 0\n78 0 0\n90 0 0\n",
"3 2 1\n3 0 1\n3 1 0\n3 0 0\n",
"27 5 1\n232 0 1\n72 0 1\n235 0 1\n2 0 1\n158 0 0\n267 0 0\n242 0 1\n1 0 0\n64 0 0\n139 1 1\n250 0 1\n208 0 1\n127 0 1\n29 0 1\n53 0 1\n100 0 1\n52 0 1\n229 0 0\n1 0 1\n29 0 0\n17 0 1\n74 0 1\n211 0 1\n248 0 1\n15 0 0\n252 0 0\n159 0 1\n",
"6 4 3\n19 0 0\n6 1 1\n57 1 0\n21 0 1\n53 1 1\n9 1 1\n"
],
"output": [
"2\n",
"2\n",
"38\n",
"6\n",
"26\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1",
"-1",
"-1\n",
"-1\n",
"-1\n",
"-1"
]
}
|
CORRECT
|
python3
|
from sys import stdin
from collections import defaultdict as dd
from collections import deque as dq
import itertools as it
from math import sqrt, log, log2
from fractions import Fraction
n, k = map(int, input().split())
alike, blike = 0, 0
zero_one, one_zero, one_one, zero_zero = [], [], [], []
for i in range(n):
t, a, b = map(int, input().split())
alike += a
blike += b
if a == 0 and b == 1:
zero_one.append((t, i))
elif a == 1 and b == 0:
one_zero.append((t, i))
elif a== 1 and b== 1:
one_one.append((t, i))
else:
pass
# zero_zero.append((t, i))
if alike < k or blike < k:
print(-1)
exit()
zero_one.sort(key = lambda x: x[0])
one_one.sort(key = lambda x: x[0])
one_zero.sort(key = lambda x: x[0])
# zero_zero.sort(key = lambda x: x[0])
alike, blike = 0, 0
zo, oo, oz, zz = 0, 0, 0, 0
lzo, loo, loz = len(zero_one), len(one_one), len(one_zero)
# lzo, loo, loz, lzz = len(zero_one), len(one_one), len(one_zero), len(zero_zero)
tottime = 0
# books = []
while alike<k or blike<k:
# if oo >= loo and zo >= lzo and oz>=loz:
# break
# lbo = len(books)
# if lbo == m:
if alike <k and blike <k:
if oo>= loo and zo>=lzo and oz>=loz:
print(-1)
exit()
elif zo >= lzo or oz >= loz:
tottime += one_one[oo][0]
# books.append(one_one[oo][1])
oo += 1
elif oo >= loo:
tottime += zero_one[zo][0] + one_zero[oz][0]
# books.append(zero_one[zo][1])
# books.append(zero_one[oz][1])
zo += 1
oz += 1
elif zero_one[zo][0] + one_zero[oz][0] < one_one[oo][0]:
tottime += zero_one[zo][0] + one_zero[oz][0]
# books.append(zero_one[zo][1])
# books.append(zero_one[oz][1])
zo += 1
oz += 1
else:
tottime += one_one[oo][0]
# books.append(one_one[oo][1])
oo += 1
alike += 1
blike += 1
elif alike == k and blike < k:
if oo>=loo and zo>=lzo:
print(-1)
exit()
elif oo >= loo:
tottime += zero_one[zo][0]
# books.append(zero_one[zo][1])
zo += 1
elif zo >= lzo:
tottime += one_one[oo][0]
# books.append(one_one[oo][1])
oo += 1
elif zero_one[zo][0] < one_one[oo][0]:
tottime += zero_one[zo][0]
# books.append(zero_one[zo][1])
zo += 1
else:
tottime += one_one[oo][0]
# books.append(one_one[oo][1])
oo += 1
blike += 1
elif alike <k and blike == k:
if oo>=loo and oz>=loz:
print(-1)
exit()
elif oo>=loo:
tottime += one_zero[oz][0]
# books.append(one_zero[oz][1])
oz += 1
elif oz>= loz:
tottime += one_one[oo][0]
# books.append(one_one[oo][1])
oo += 1
elif one_zero[oz][0] < one_one[oo][0]:
tottime += one_zero[oz][0]
# books.append(one_zero[oz][1])
oz += 1
else:
tottime += one_one[oo][0]
# books.append(one_one[oo][1])
oo += 1
print(tottime)
# print(*books)
|
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