Search is not available for this dataset
name 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 983k |
|---|---|---|---|---|---|---|
1012_E. Cycle sort | You are given an array of n positive integers a_1, a_2, ..., a_n. You can perform the following operation any number of times: select several distinct indices i_1, i_2, ..., i_k (1 ≤ i_j ≤ n) and move the number standing at the position i_1 to the position i_2, the number at the position i_2 to the position i_3, ..., t... | {
"input": [
"5 5\n3 2 3 1 1\n",
"4 3\n2 1 4 3\n",
"2 0\n2 2\n"
],
"output": [
"1\n5\n1 4 2 3 5 \n",
"-1\n",
"0\n"
]
} | {
"input": [
"5 0\n884430748 884430748 708433020 708433020 708433020\n",
"2 1\n1 1\n",
"2 0\n2 1\n",
"5 2\n65390026 770505072 65390026 65390026 65390026\n",
"5 4\n812067558 674124159 106041640 106041640 674124159\n",
"5 4\n167616600 574805150 651016425 150949603 379708534\n",
"5 5\n4727783... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using VI = vector<int>;
const int NN = 200011;
int a[NN], b[NN], arr[NN];
VI vec[NN], cyc[NN];
int nc;
int dp[NN], vst[NN];
int n, s;
void dfs(int u) {
if (vec[u].empty()) return;
int v = vec[u].back();
vec[u].pop_back();
dfs(a[v]);
cyc[nc].push_back(v);
}
int sol... |
1012_E. Cycle sort | You are given an array of n positive integers a_1, a_2, ..., a_n. You can perform the following operation any number of times: select several distinct indices i_1, i_2, ..., i_k (1 ≤ i_j ≤ n) and move the number standing at the position i_1 to the position i_2, the number at the position i_2 to the position i_3, ..., t... | {
"input": [
"5 5\n3 2 3 1 1\n",
"4 3\n2 1 4 3\n",
"2 0\n2 2\n"
],
"output": [
"1\n5\n1 4 2 3 5 \n",
"-1\n",
"0\n"
]
} | {
"input": [
"5 0\n884430748 884430748 708433020 708433020 708433020\n",
"2 1\n1 1\n",
"2 0\n2 1\n",
"5 2\n65390026 770505072 65390026 65390026 65390026\n",
"5 4\n812067558 674124159 106041640 106041640 674124159\n",
"5 4\n167616600 574805150 651016425 150949603 379708534\n",
"5 5\n4727783... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <class T, class U>
void ckmin(T &a, U b) {
if (a > b) a = b;
}
template <class T, class U>
void ckmax(T &a, U b) {
if (a < b) a = b;
}
const int MAXN = 400013;
int N, S, M, ans, n;
int val[MAXN], arr[MAXN], sorted[MAXN];
vector<int> moves[MAXN];
vector<int> cyc... |
1012_E. Cycle sort | You are given an array of n positive integers a_1, a_2, ..., a_n. You can perform the following operation any number of times: select several distinct indices i_1, i_2, ..., i_k (1 ≤ i_j ≤ n) and move the number standing at the position i_1 to the position i_2, the number at the position i_2 to the position i_3, ..., t... | {
"input": [
"5 5\n3 2 3 1 1\n",
"4 3\n2 1 4 3\n",
"2 0\n2 2\n"
],
"output": [
"1\n5\n1 4 2 3 5 \n",
"-1\n",
"0\n"
]
} | {
"input": [
"5 0\n884430748 884430748 708433020 708433020 708433020\n",
"2 1\n1 1\n",
"2 0\n2 1\n",
"5 2\n65390026 770505072 65390026 65390026 65390026\n",
"5 4\n812067558 674124159 106041640 106041640 674124159\n",
"5 4\n167616600 574805150 651016425 150949603 379708534\n",
"5 5\n4727783... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <typename T>
vector<T>& operator--(vector<T>& v) {
for (auto& i : v) --i;
return v;
}
template <typename T>
vector<T>& operator++(vector<T>& v) {
for (auto& i : v) ++i;
return v;
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
for... |
1012_E. Cycle sort | You are given an array of n positive integers a_1, a_2, ..., a_n. You can perform the following operation any number of times: select several distinct indices i_1, i_2, ..., i_k (1 ≤ i_j ≤ n) and move the number standing at the position i_1 to the position i_2, the number at the position i_2 to the position i_3, ..., t... | {
"input": [
"5 5\n3 2 3 1 1\n",
"4 3\n2 1 4 3\n",
"2 0\n2 2\n"
],
"output": [
"1\n5\n1 4 2 3 5 \n",
"-1\n",
"0\n"
]
} | {
"input": [
"5 0\n884430748 884430748 708433020 708433020 708433020\n",
"2 1\n1 1\n",
"2 0\n2 1\n",
"5 2\n65390026 770505072 65390026 65390026 65390026\n",
"5 4\n812067558 674124159 106041640 106041640 674124159\n",
"5 4\n167616600 574805150 651016425 150949603 379708534\n",
"5 5\n4727783... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 100;
int n, limit;
int nxt[N], lab[N], flag[N];
pair<int, int> a[N];
vector<int> ans1, ans2;
vector<vector<int> > ans3;
vector<pair<int, int> > v;
int root(int u) {
if (lab[u] < 0) return u;
return lab[u] = root(lab[u]);
}
void join(int u, int v) {
... |
1012_E. Cycle sort | You are given an array of n positive integers a_1, a_2, ..., a_n. You can perform the following operation any number of times: select several distinct indices i_1, i_2, ..., i_k (1 ≤ i_j ≤ n) and move the number standing at the position i_1 to the position i_2, the number at the position i_2 to the position i_3, ..., t... | {
"input": [
"5 5\n3 2 3 1 1\n",
"4 3\n2 1 4 3\n",
"2 0\n2 2\n"
],
"output": [
"1\n5\n1 4 2 3 5 \n",
"-1\n",
"0\n"
]
} | {
"input": [
"5 0\n884430748 884430748 708433020 708433020 708433020\n",
"2 1\n1 1\n",
"2 0\n2 1\n",
"5 2\n65390026 770505072 65390026 65390026 65390026\n",
"5 4\n812067558 674124159 106041640 106041640 674124159\n",
"5 4\n167616600 574805150 651016425 150949603 379708534\n",
"5 5\n4727783... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, k, tot, x, ans, a[1000000], b[1000000], f[1000000], ed[1000000],
nt[1000000], nm[1000000], ss[1000000];
map<int, int> mp, rw;
bool v[1000000];
int gf(int x) {
if (f[x] == x) return x;
f[x] = gf(f[x]);
return f[x];
}
int main() {
scanf("%d %d", &n, &k);
... |
1012_E. Cycle sort | You are given an array of n positive integers a_1, a_2, ..., a_n. You can perform the following operation any number of times: select several distinct indices i_1, i_2, ..., i_k (1 ≤ i_j ≤ n) and move the number standing at the position i_1 to the position i_2, the number at the position i_2 to the position i_3, ..., t... | {
"input": [
"5 5\n3 2 3 1 1\n",
"4 3\n2 1 4 3\n",
"2 0\n2 2\n"
],
"output": [
"1\n5\n1 4 2 3 5 \n",
"-1\n",
"0\n"
]
} | {
"input": [
"5 0\n884430748 884430748 708433020 708433020 708433020\n",
"2 1\n1 1\n",
"2 0\n2 1\n",
"5 2\n65390026 770505072 65390026 65390026 65390026\n",
"5 4\n812067558 674124159 106041640 106041640 674124159\n",
"5 4\n167616600 574805150 651016425 150949603 379708534\n",
"5 5\n4727783... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 1;
int a[N], p[N], head[N], now = 0;
pair<int, int> b[N];
vector<int> cycle[N];
int findd(int x) {
if (head[x] < 0) {
return x;
}
int y = findd(head[x]);
head[x] = y;
return y;
}
void unionn(int x, int y) {
x = findd(x), y = findd(y);
h... |
1012_E. Cycle sort | You are given an array of n positive integers a_1, a_2, ..., a_n. You can perform the following operation any number of times: select several distinct indices i_1, i_2, ..., i_k (1 ≤ i_j ≤ n) and move the number standing at the position i_1 to the position i_2, the number at the position i_2 to the position i_3, ..., t... | {
"input": [
"5 5\n3 2 3 1 1\n",
"4 3\n2 1 4 3\n",
"2 0\n2 2\n"
],
"output": [
"1\n5\n1 4 2 3 5 \n",
"-1\n",
"0\n"
]
} | {
"input": [
"5 0\n884430748 884430748 708433020 708433020 708433020\n",
"2 1\n1 1\n",
"2 0\n2 1\n",
"5 2\n65390026 770505072 65390026 65390026 65390026\n",
"5 4\n812067558 674124159 106041640 106041640 674124159\n",
"5 4\n167616600 574805150 651016425 150949603 379708534\n",
"5 5\n4727783... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct ln {
int val;
ln* next;
};
vector<pair<int, int> >* nl;
vector<pair<int, int> > ar;
int* roi;
bool* btdt;
ln* fe(ln* ath, int cp) {
ln* sv = NULL;
for (int i = roi[cp]; i < nl[cp].size(); i = roi[cp]) {
int np = nl[cp][i].first;
int ind = nl[cp][i].se... |
1012_E. Cycle sort | You are given an array of n positive integers a_1, a_2, ..., a_n. You can perform the following operation any number of times: select several distinct indices i_1, i_2, ..., i_k (1 ≤ i_j ≤ n) and move the number standing at the position i_1 to the position i_2, the number at the position i_2 to the position i_3, ..., t... | {
"input": [
"5 5\n3 2 3 1 1\n",
"4 3\n2 1 4 3\n",
"2 0\n2 2\n"
],
"output": [
"1\n5\n1 4 2 3 5 \n",
"-1\n",
"0\n"
]
} | {
"input": [
"5 0\n884430748 884430748 708433020 708433020 708433020\n",
"2 1\n1 1\n",
"2 0\n2 1\n",
"5 2\n65390026 770505072 65390026 65390026 65390026\n",
"5 4\n812067558 674124159 106041640 106041640 674124159\n",
"5 4\n167616600 574805150 651016425 150949603 379708534\n",
"5 5\n4727783... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 202020, d = 0;
int N, S, oval[MAXN], edge[MAXN], comp[MAXN], cnum = 0, vis[MAXN], vis2[MAXN];
vector<int> comp_stack[MAXN];
pair<int, int> vals[MAXN];
vector<int> cursol;
vector<vector<int> > solution;
void recur(int n) {
if (vis[n] > 0) return;
vis[n] ... |
1012_E. Cycle sort | You are given an array of n positive integers a_1, a_2, ..., a_n. You can perform the following operation any number of times: select several distinct indices i_1, i_2, ..., i_k (1 ≤ i_j ≤ n) and move the number standing at the position i_1 to the position i_2, the number at the position i_2 to the position i_3, ..., t... | {
"input": [
"5 5\n3 2 3 1 1\n",
"4 3\n2 1 4 3\n",
"2 0\n2 2\n"
],
"output": [
"1\n5\n1 4 2 3 5 \n",
"-1\n",
"0\n"
]
} | {
"input": [
"5 0\n884430748 884430748 708433020 708433020 708433020\n",
"2 1\n1 1\n",
"2 0\n2 1\n",
"5 2\n65390026 770505072 65390026 65390026 65390026\n",
"5 4\n812067558 674124159 106041640 106041640 674124159\n",
"5 4\n167616600 574805150 651016425 150949603 379708534\n",
"5 5\n4727783... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, s, a[200010];
map<int, int> mp1;
map<int, vector<int> > mp2;
vector<vector<int> > ans;
void dfs(int u) {
while (!mp2[u].empty()) {
int i = mp2[u].back();
mp2[u].pop_back();
dfs(a[i]);
ans.back().push_back(i);
}
mp2.erase(u);
}
int main() {
sca... |
1012_E. Cycle sort | You are given an array of n positive integers a_1, a_2, ..., a_n. You can perform the following operation any number of times: select several distinct indices i_1, i_2, ..., i_k (1 ≤ i_j ≤ n) and move the number standing at the position i_1 to the position i_2, the number at the position i_2 to the position i_3, ..., t... | {
"input": [
"5 5\n3 2 3 1 1\n",
"4 3\n2 1 4 3\n",
"2 0\n2 2\n"
],
"output": [
"1\n5\n1 4 2 3 5 \n",
"-1\n",
"0\n"
]
} | {
"input": [
"5 0\n884430748 884430748 708433020 708433020 708433020\n",
"2 1\n1 1\n",
"2 0\n2 1\n",
"5 2\n65390026 770505072 65390026 65390026 65390026\n",
"5 4\n812067558 674124159 106041640 106041640 674124159\n",
"5 4\n167616600 574805150 651016425 150949603 379708534\n",
"5 5\n4727783... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct ln {
int val;
ln* next;
};
vector<pair<int, int> >* nl;
int* roi;
bool* btdt;
ln* fe(ln* ath, int cp) {
for (int i = roi[cp]; i < nl[cp].size(); i = roi[cp]) {
int np = nl[cp][i].first;
int ind = nl[cp][i].second;
if (btdt[ind]) {
continue;
... |
1012_E. Cycle sort | You are given an array of n positive integers a_1, a_2, ..., a_n. You can perform the following operation any number of times: select several distinct indices i_1, i_2, ..., i_k (1 ≤ i_j ≤ n) and move the number standing at the position i_1 to the position i_2, the number at the position i_2 to the position i_3, ..., t... | {
"input": [
"5 5\n3 2 3 1 1\n",
"4 3\n2 1 4 3\n",
"2 0\n2 2\n"
],
"output": [
"1\n5\n1 4 2 3 5 \n",
"-1\n",
"0\n"
]
} | {
"input": [
"5 0\n884430748 884430748 708433020 708433020 708433020\n",
"2 1\n1 1\n",
"2 0\n2 1\n",
"5 2\n65390026 770505072 65390026 65390026 65390026\n",
"5 4\n812067558 674124159 106041640 106041640 674124159\n",
"5 4\n167616600 574805150 651016425 150949603 379708534\n",
"5 5\n4727783... | IN-CORRECT | cpp | #include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <functional>
#include <iostream>
#include <map>
#include <memory>
#include <numeric>
#include <queue>
#include <random>
#incl... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int module = 1000000007;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n, m, k;
cin >> n >> m >> k;
vector<set<int>> to(n, set<int>());
vector<pair<int, int>> edges(m);
vector<int> nbr(n);
for (int i = 0; i < m; ++i) {
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
#pragma GCC optimize(3, "Ofast", "inline")
using namespace std;
const long long maxn = (long long)1e6 + 5;
const long long mod = (long long)998244353;
const long long inf = 0x3f3f3f3f3f3f3f3f;
pair<long long, long long> a[maxn];
long long in[maxn];
vector<long long> v[maxn];
bool vis[maxn];
lon... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long N = 3e5 + 10;
long long x[N], y[N], deg[N];
vector<long long> vp[N];
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
long long n, m, k;
cin >> n >> m >> k;
for (long long i = 1; i <= m; i++) {
cin >... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.io.*;
import java.util.*;
public class Test {
static int readInt() {
int ans = 0;
boolean neg = false;
try {
boolean start = false;
for (int c = 0; (c = System.in.read()) != -1; ) {
if (c == '-') {
start = true;
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.util.*;
import java.io.*;
public class P1037E {
private static void solve() {
int n = nextInt();
int m = nextInt();
int k = nextInt();
Map<Integer, Set<Integer>> friends = new HashMap<>();
int[] f = new int[m];
int[] t = new int[m];
for (int i = 0; i < m; i++) {
int ... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
set<int> se[200100];
int u[200100];
int v[200100];
int vis[200100];
int n, m, k, ans;
int num[200100];
void dfs(int st) {
if (se[st].size() >= k || vis[st]) return;
vis[st] = 1;
--ans;
for (auto &t : se[st]) {
se[t].erase(st);
dfs(t);
}
se[st].clear();
}... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
inline int read() {
int sum = 0, ff = 1;
char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') ff = -1;
ch = getchar();
}
while (isdigit(ch)) sum = sum * 10 + (ch ^ 48), ch = getchar();
return sum * ff;
}
const int mod = 1e9 + 7;
const int mo = 9982... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 200100;
struct node {
int v, u;
} e[maxn + 10];
set<int> s[maxn];
int k, ans, in[maxn], vis[maxn], ret[maxn], n, m;
queue<int> q;
void dele(int x) {
set<int>::iterator iter;
int u, v;
if (vis[x] || in[x] >= k) return;
vis[x] = 1;
q.push(x);
--... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int mx = 200 * 1000 + 5;
vector<set<int> > vec(mx);
int arr1[mx], arr2[mx];
int main() {
int i, j, k, n, m, x, y, p, q, l;
scanf("%d %d %d", &n, &m, &k);
int ans = 0;
for (i = 0; i < m; i++) {
scanf("%d %d", &x, &y);
arr1[i] = x;
arr2[i] = y;
v... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
const int ms = 200200;
int n, m, k;
int ans[ms];
std::pair<int, int> edges[ms];
std::vector<int> use[ms];
int deg[ms];
bool del[ms];
int main() {
std::cin >> n >> m >> k;
for (int i = 0; i < m; i++) {
int u, v;
scanf("%d %d", &u, &v);
use[u].push_back(i);
use[v].push_back(i)... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.OutputStream;
import java.util.Arrays;
import java.util.Iterator;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.FileNotFoundException;
import java.util.StringTok... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 100;
vector<pair<int, int>> g[N];
vector<int> deg, ans;
vector<pair<int, int>> edge;
set<pair<int, int>> good;
vector<bool> in_good;
void f(int k, int iteration) {
while (!good.empty() && good.begin()->first < k) {
int v = good.begin()->second;
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
const double pi = acos(-1.0);
const int maxn = 200000 + 10;
const int mod = 1e9 + 7;
inline char _getchar() {
static const int BUFSIZE = 100001;
static char buf[BUFSIZE];
static char *psta = buf, ... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using namespace std;
int deg[200100];
set<pair<int, int> > pset;
vector<pair<int, int> > query;
set<int> conn[200100];
bool is_going(int u) { return (pset.find(make_pair(deg[u], u)) != pset.end()); }
int n, m, k;
void delete_adjacent(int u) {
for (auto v : conn[u]) {
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOExcept... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int INF = 2147483647;
const long long INFL = 9223372036854775807LL;
const double EPSILON = 0.00000001;
const long long MOD = 1000000007;
vector<int> adj[200000 + 5];
unordered_map<int, int> removed[200000 + 5];
int degree[200000 + 5];
bool node_removed[200000 + 5];
in... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.io.*;
import java.util.*;
import javafx.util.Pair;
public class HR27 {
InputStream is;
PrintWriter out;
String INPUT = "";
//boolean codechef=true;
boolean codechef=true;
void solve()
{
int n=ni(),m=ni(),k=ni();
int[] from=new int[2*m];
Pair[] to=new Pair[2*m];
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
int n, m, k, u[200010], v[200010], t[200010];
std::set<int> next[200010], p;
inline void remove(int x) {
if ((int)next[x].size() < k && p.erase(x)) {
for (auto y : next[x]) {
next[y].erase(x);
remove(y);
}
next[x].clear();
}
}
int main() {
scanf("%d%d%d", &n, &m, &... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.io.*;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;
import java.util.StringTokenizer;
public class E {
FastScanner in;
PrintWriter out;
public static void main(String[] args) {
new E().run();
}
void solve() throws IOException {
int n = in.nex... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.io.*;
import java.math.*;
import java.util.*;
import java.util.stream.*;
public class E {
void submit() {
int n = nextInt();
int m = nextInt();
int k = nextInt();
int[] vs = new int[m];
int[] us = new int[m];
int[] deg = new int[n];
for (int i = 0; i < m; i++) {
vs[i] = nextInt() - 1;... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m, k;
set<int> M[200005];
int siz[200005], a[200005], b[200005], dele[200005], cnt, ans[200005];
void check(int u) {
if (siz[u] >= k || dele[u]) return;
cnt--;
dele[u] = 1;
queue<int> Q;
Q.push(u);
while (!Q.empty()) {
int ty = Q.front();
Q.pop();... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.io.*;
import java.lang.reflect.Field;
import java.util.*;
public class Main {
long MOD = (long) (1e9 + 7);
HashSet<Integer>[] graph;
static int[] deg;
int rem;
public static void main(String[] args) throws IOException {
Reader.init(System.in);
new Main();
}
M... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int du[maxn];
set<int> g[maxn];
int b1[maxn], b2[maxn], ans[maxn], vis[maxn];
int main() {
int n, m, k;
scanf("%d%d%d", &n, &m, &k);
set<pair<int, int> > s;
for (int i = 0; i < m; i++) {
int x, y;
scanf("%d%d", &x, &y);
b1[i] =... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
/**
*
*/
public class E {
static LinkedList<Pair>[] g;
static int k;
static boolean[] s;
static int size;
static int[] degree;
static Queue<Integer> q = new LinkedList<>();
p... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
template <typename T, typename T2>
inline void _max(T &a, T2 b) {
a = max((T)a, (T)b);
}
template <typename T, typename T2>
inline void _min(T &a, T2 b) {
a = min((T)a, (T)b);
}
const int MAX = 2e5 +... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int mx = 2e5 + 10, cut = 700;
int n, m, k, cnt[mx], lv, x[mx], y[mx], dead[mx];
vector<pair<int, int> > adj[mx];
vector<int> res;
void er(int a, int b) {
auto it = lower_bound(adj[a].begin(), adj[a].end(), make_pair(b, 0));
if (it->second) return;
it->second = 1... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.Arrays;
import java.io.FilterInputStream;
import java.io.BufferedInputStream;
import java.util.TreeSet;
import java.io.InputStream;
/**
* @author khokharnikunj8
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > adj;
set<long long> st;
vector<long long> deg;
long long k;
set<pair<long long, long long> > edges;
void remove(long long r) {
st.erase(r);
for (long long i = 0; i < adj[r].size(); i++) {
long long c = adj[r][i];
if (st.find(c) == st.e... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int n, m, k;
bool vis[N];
queue<int> Q;
set<int> G[N];
vector<int> E[N];
int a[N], b[N], ans[N], sum;
void solve() {
while (!Q.empty()) {
int u = Q.front();
Q.pop();
for (auto v : G[u]) {
G[v].erase(u);
if (!vis[v] && G[v].s... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main(int agrc, char* argv[]) {
int n, m, k;
scanf("%d %d %d", &n, &m, &k);
std::vector<std::vector<int>> edges(n + 1);
int selected[n];
for (int i = 0; i < n; ++i) {
selected[i] = 1;
}
int total = n;
int friends[m][2];
for (int i = 0; i < m; ++i) {... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.io.*;
import java.util.*;
public class A {
public static void main (String[] args) { new A(); }
int curSize;
boolean[] killed;
A() {
FastScanner s = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
System.err.println("");
int n = s.nextInt();
int m = s.nextInt();
int k ... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
cout << fixed << setprecision(15);
const int LIM = 600;
int N, M, K;
cin >> N >> M >> K;
vector<int> x(M), y(M);
vector<vector<int> > g(N);
for (int i = 0; i < M; i++) {
cin >> x[i... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
#pragma GCC optimize(3, "Ofast", "inline")
using namespace std;
bool Finish_read;
template <class T>
inline void read(T &x) {
Finish_read = 0;
x = 0;
int f = 1;
char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') f = -1;
if (ch == EOF) return;
ch = getchar();
}
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.io.*;
import java.util.*;
public class Mainn {
FastReader scn;
PrintWriter out;
String INPUT = "";
void solve() {
int n = scn.nextInt(), m = scn.nextInt(), k = scn.nextInt();
int[] from = new int[m], to = new int[m], w = new int[m];
for (int i = 0; i < m; i++) {
from[i] = scn.nextInt() - 1;
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.util.*;
import java.io.*;
public class Main {
public static void main(String args[]) {new Main().run();}
FastReader in = new FastReader();
PrintWriter out = new PrintWriter(System.out);
void run(){
work();
out.flush();
}
long mod=1000000007;
long gcd(long a,long b) {
return b==0?a:gcd(b,a%b);... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | /**
* BaZ :D
*/
import java.util.*;
import java.io.*;
import static java.lang.Math.*;
public class Main
{
static Reader scan;
static PrintWriter pw;
static HashSet<Integer> good;
static HashSet<Integer> adj[];
static int n,m,k,ans[];
public static void main(String[] args) {
new Threa... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m, k, numOn, res[200000];
set<int> f[200000];
bool on[200000];
vector<pair<int, int> > v;
queue<int> toRemove;
void update() {
while (!toRemove.empty()) {
int r = toRemove.front();
toRemove.pop();
numOn--;
for (auto next : f[r]) {
f[next].eras... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.OutputStream;
import java.util.Arrays;
import java.util.Iterator;
import java.io.IOException;
import java.util.AbstractSequentialList;
import java.io.InputStreamReader;
import java.util.TreeSe... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long N = 5e5;
const long long INF = 1e18;
const long long mod = 1e9 + 7;
set<long long> g[N], st;
long long i, j, n, m, k, ans[N], u[N], v[N];
void dfs(long long x) {
auto it = st.lower_bound(x);
if (*it != x || g[x].size() >= k) return;
st.erase(it);
for... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.io.*;
import java.util.*;
import java.util.stream.IntStream;
public class E2 {
public static void main(String[] args) throws IOException {
E2 runner = new E2();
runner.run();
runner.close();
}
BufferedReader input = new BufferedReader(new InputStreamReader(System.in));... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | //package sumo;
import java.io.*;
import java.util.*;
public class Tester {
static TreeSet<Integer> tr[];
static int k,n,m,cnt=0;
static Queue<Integer> q;
static TreeSet<Integer>glo;
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
FastReader s=new FastReader();
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | python2 |
import sys
range = xrange
n,m,k = [int(x) for x in sys.stdin.readline().split()]
inp = [int(x)-1 for x in sys.stdin.read().split()]
ii = 0
coupl = [[] for _ in range(n)]
time = [[] for _ in range(n)]
nfr = [0]*n
for i in range(m):
a,b = inp[ii],inp[ii+1]
ii += 2
coupl[a].append(b)
coupl[b].append(a... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.io.*;
import java.util.*;
public class Trips {
public static void main(String[] args) {
FastScanner scan=new FastScanner();
PrintWriter out=new PrintWriter(System.out);
int n=scan.nextInt(), m=scan.nextInt();
k=scan.nextInt();
deg=new int[n];
a=new ArrayList[n];
for(int i=0;i<n;i++) a[i]=new... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.Stack;
import java.util.TreeSet;
/**
* Created by himanshubhardwaj on 04/09/18.
*/
public class AllVertecesWithMinimimKRemain... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T& x) {
char c = getchar();
bool f = false;
for (x = 0; !isdigit(c); c = getchar()) {
if (c == '-') {
f = true;
}
}
for (; isdigit(c); c = getchar()) {
x = x * 10 + c - '0';
}
if (f) {
x = -x;
}
}
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
vector<pair<int, int>> edges, g[N];
vector<int> ans;
set<pair<int, int>> s;
set<int> deleted;
int n, m, k, a, b, deg[N];
int main() {
scanf("%d%d%d", &n, &m, &k);
for (int i = 0; i < m; ++i) {
scanf("%d%d", &a, &b);
edges.emplace_back(a, ... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long N = 3e5 + 10;
long long x[N], y[N], deg[N];
vector<long long> vp[N];
long long removed[N];
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
long long n, m, k;
cin >> n >> m >> k;
for (long long i = 1; i ... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, k;
int res;
vector<set<int>> g;
vector<int> used;
vector<int> selected;
void bfs(int start) {
vector<int> seen = {start};
queue<int> q;
used[start] = 1;
q.push(start);
while (!q.empty()) {
int v = q.front();
q.pop();
assert(selected[v]);
ass... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.io.InputStream;
/**
* Bui... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
int n, m, k, deg[N], ans[N];
pair<int, int> edges[N];
set<int> adj[N];
set<pair<int, int> > st;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> k;
for (int i = 1; i <= m; i++) {
int x, y;
cin >> x... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
pair<int, int> edge[maxn];
map<pair<int, int>, int> mp;
vector<int> g[maxn];
int vis[maxn];
queue<int> q;
int du[maxn];
int ans[maxn];
int main() {
std::ios::sync_with_stdio(false);
int n, m, k;
cin >> n >> m >> k;
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, m, k;
cin >> n >> m >> k;
vector<vector<pair<int, int>>> g(n);
vector<int> from(m), to(m);
for (int i = 0; i < m; i++) {
cin >> from[i] >> to[i];
from[i]--;
to[i]--;
g[from[i]].emplac... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = int(200000) + 5;
int t, n, m, q, k, cnt, sum, x, y, l, r;
int a[N], b[N], ans[N];
pair<int, int> edge[N];
vector<pair<int, int> > v[N];
set<pair<int, int> > s;
set<pair<int, int> >::iterator it;
void update(int I) {
while (!s.empty() && (s.begin()->first) < ... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using LL = long long;
constexpr int N = 2e5 + 5;
int ans[N];
int degree[N];
int is_good[N];
vector<pair<int, int>> E[N];
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, m, k;
cin >> n >> m >> k;
vector<pair<int, int>> e(m);
for (int i = 0; i <... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m, k, x[200009], y[200009];
set<int> alive;
vector<int> kaaj;
set<int> graph[200009];
int degree[200009], ans[200009];
void doit(int z) {
if (alive.find(z) == alive.end()) return;
alive.erase(z);
if (graph[z].empty()) return;
auto it = graph[z].begin();
whi... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.util.*;
import java.io.*;
import java.text.*;
//Solution Credits: Taranpreet Singh
public class Main{
//SOLUTION BEGIN
void solve(int TC) throws Exception{
int n = ni(), m = ni(), k = ni();
TreeSet<Long> set = new TreeSet<>();
HashSet<Integer>[] g = new HashSet[n];
f... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.util.*;
import java.lang.*;
import java.math.*;
import java.io.*;
import static java.lang.Math.*;
/* spar5h */
//hm.get((long)0) != hm.get(0)
public class cf5 implements Runnable{
static class pair {
int i, j;
pair(int i, int j) {
this.i = i; this.j = j;
}
}
public void run() {
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
set<int> adj[200009];
set<int>::iterator it;
int vis[200009];
pair<int, int> edg[200009];
vector<int> bad;
int ses[200009];
int main() {
int n, m, k;
cin >> n >> m >> k;
for (int i = 1; i <= m; i++) {
scanf("%d%d", &edg[i].first, &edg[i].second);
adj[edg[i].fi... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct Edge {
int x, y;
} E[200010];
struct edge {
int nxt, t, s;
} e[200010 << 1];
int head[200010], edge_cnt;
void add_edge(int x, int y, int z) {
e[edge_cnt] = (edge){head[x], y, z};
head[x] = edge_cnt++;
}
bool ban[200010];
int Deg[200010], Q[200010], Ans[200010... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int intcmp(const void *v1, const void *v2) { return *(int *)v1 - *(int *)v2; }
int n, m, k;
vector<pair<int, int>> v1v2index[200010];
int ea[200010], eb[200010];
int d[200010];
int curidx;
bool deleted[200010];
int delcnt[200010];
int delnow;
void dfs(int v) {
if (deleted... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
const int N = 2e5 + 5;
using namespace std;
int n, m, k;
int res[N];
int x[N], y[N];
int bac[N];
set<int> a[N];
set<pair<int, int> > S;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> k;
for (int i = 1; i <= m; i++) {
cin >> x[i] >> y[i];
a... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int n, m, k;
cin >> n >> m >> k;
vector<int> from(m), to(m);
vector<vector<pair<int, int> > > adj(n);
for (int i = (int)(0); i < (int)(m); ++i) {
cin >> from[i] >> to[i];
from[i]--, to[i... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.util.Set;
import java.util.InputMismatchException;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.io.Writer;
import java.i... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
set<int> S, G[N];
int A[N], B[N], C[N], n, m, k;
void U(int u) {
if (G[u].size() < k && S.erase(u))
for (auto v : G[u]) G[v].erase(u), U(v);
}
int main() {
cin >> n >> m >> k;
for (int i = 1; i <= m; ++i) {
scanf("%d%d", A + i, B + i);
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int MAX = 2e5 + 1;
int n, m, k, ans;
int du[MAX], X[MAX], Y[MAX], Ans[MAX];
vector<int> ve[MAX];
bool use[MAX];
void work(int x) {
queue<int> qu;
qu.push(x), use[x] = 1, --ans;
while (!qu.empty()) {
int tt = qu.front();
qu.pop();
for (int i = 0; i < ... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long mod = 1000000007;
const double e = 2.718281828459;
long long gcd(long long a, long long b) {
if (!b) return a;
return gcd(b, a % b);
}
long long gcd1(int a, int b) {
if (!b) return a;
return gcd(b, a % b);
}
long long pow_... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int x[200010], y[200010];
set<int> s[200010];
queue<int> q, p;
bool inq[200010];
bool w[200010];
int ans[200010];
int n, m, k, sum;
void reset() {
while (!q.empty()) q.pop();
while (!p.empty()) inq[p.front()] = false, p.pop();
}
void BFS() {
while (!q.empty()) {
i... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long linf = 1LL << 62;
const int iinf = 1000000009;
const double dinf = 1e17;
const int Mod = 1e9 + 9;
const int maxn = 1000005;
set<pair<int, int> > s, sp;
int n, m, k, u, v;
int mp[maxn], rt;
pair<int, int> sd[maxn], ds[maxn];
vector<int> vp[200005];
void up(in... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int OO = 0x3f3f3f3f;
const double eps = (1e-10);
stringstream out;
int n, m, k, u, v, deg[200005];
bool isGood[200005];
vector<int> ans(200005);
set<pair<int, int> > good;
vector<pair<int, int> > edges(200005);
vector<vector<pair<int, int> > > adj(200005);
int main() ... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.util.*;
import java.io.*;
public class E18 {
public static void main(String[] args) {
MyScanner sc = new MyScanner();
PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
int n = sc.nextInt(); int m = sc.nextInt(); int k = sc.nextInt();
HashSet<Intege... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m, cnt, k, tot, tp;
int head[200005];
int sta[200005];
int dgr[200005];
int ans[200005];
bool usd[200005];
struct node {
int fr;
int to;
int nxt;
int mrk;
} edge[400005];
void init() {
cnt = 0;
memset(head, -1, sizeof(head));
}
void addedge(int f, int t) ... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using lint = long long;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-16;
int N, M, K;
int edges[200010][2], degree[200010];
set<int> graph[200010];
set<tuple<int, int>> nodes;
int ans[200010];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout << setprec... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | import java.io.*;
import java.math.*;
import java.util.*;
import static java.util.Arrays.fill;
import static java.lang.Math.*;
import static java.util.Arrays.sort;
import static java.util.Collections.sort;
public class Trips {
static int mod = 1000000007;
static InputReader in = new InputReader(System.in);
static ... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | /* Author: Ronak Agarwal, reader part not written by me*/
import java.io.* ; import java.util.* ;
import static java.lang.Math.min ; import static java.lang.Math.max ;
import static java.lang.Math.abs ; import static java.lang.Math.log ;
import static java.lang.Math.pow ; import static java.lang.Math.sqrt ;
/* Thread i... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <class u>
pair<u, u> ordered_pair(const u& a, const u& b) {
return a < b ? make_pair(a, b) : make_pair(b, a);
}
int n, m, k;
set<int> gr[201010];
vector<pair<int, int>> edge_list;
set<pair<int, int>> removed_edges;
struct cmp_deg {
bool operator()(int u, int v)... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
unordered_set<int> e[N];
int n, m, k, ex[N], ey[N], i, ans, ot[N];
bool bb[N];
void del(int x) {
bb[x] = 0;
--ans;
for (int y : e[x]) e[y].erase(x);
for (int y : e[x])
if (bb[y] && e[y].size() < k) del(y);
e[x].clear();
}
int main() {
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <class T>
inline T sqr(T x) {
return x * x;
}
template <class T>
inline T parse(const string &s) {
T x;
stringstream ss(s);
ss >> x;
return x;
}
const double EPS = 1e-12;
const int INF = 1000 * 1000 * 1000;
const long long LINF = INF * 1ll * INF;
const do... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 200;
int n, m, k, id[N], vis[N], ans[N];
vector<pair<int, int> > g[N];
struct node {
int x, y;
} e[N];
set<pair<int, int> > s;
void change(int x) {
if (vis[x]) return;
auto it = s.lower_bound(make_pair(id[x], x));
pair<int, int> t = *it;
s.eras... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
int n, m, k, ans;
std::set<int> g[200005];
std::vector<std::pair<int, int> > qr;
bool deleted[200005];
void check(int v) {
if (deleted[v] || g[v].size() >= k) return;
deleted[v] = true;
ans--;
for (auto &i : g[v]) g[i].erase(v);
for (auto &i : g[v]) check(i);
}
int main() {
std::ios... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
const int N = 2e5 + 5;
int n, m, k;
vector<pair<int, int>> edge;
set<int> adj[N];
bool gone[N];
vector<int> res;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> k;
for (int i = 1; i... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long binpow(long long a, long long n, long long mod) {
long long ans = 1;
while (n) {
if (n & 1) ans = (ans * a) % mod;
a = (a * a) % mod;
n >>= 1;
}
return ans;
}
long long binpow(long long a, long long n) {
long long ans = 1;
while (n) {
i... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | java | //package baobab;
import java.io.*;
import java.util.*;
public class E {
public static void main(String[] args) {
Solver solver = new Solver();
}
static class Solver {
IO io;
public Solver() {
this.io = new IO();
try {
solve();
... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
pair<int, int> edge[200000 + 324];
vector<int> adj[200000 + 324];
set<int> degree[200000 + 324];
set<int> friends;
int ans[200000 + 324];
int k;
void rem(int x) {
set<int> elem_to_remove;
for (auto bc : degree[x]) {
elem_to_remove.insert(bc);
degree[bc].erase(x)... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxN = 1e6 + 10;
priority_queue<int> P;
priority_queue<int, vector<int>, greater<int> > mP;
long long gcd(long long a, long long b) {
if (b == 0)
return a;
else
return gcd(b, a % b);
}
int mod = 1000 * 1000 * 1000 + 7;
long long ksm(int a, int b) {
i... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m, k;
set<pair<int, int>> S;
vector<int> A;
vector<pair<int, int>> R;
vector<vector<int>> G;
set<pair<int, int>> B;
int main() {
cin >> n >> m >> k;
R.assign(m, {0, 0});
A.assign(n, 0);
G.assign(n, vector<int>(0, 0));
for (auto &i : R) {
cin >> i.first ... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main(int argc, char const *argv[]) {
int n, m, k;
cin >> n >> m >> k;
vector<vector<int> > edges(m, vector<int>(2));
vector<vector<int> > graph(n + 1);
vector<int> degrees(n + 1, 0);
for (int i = 0; i < m; ++i) {
scanf("%d %d", &edges[i][0], &edges[i][1]... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m, k, x[200001], y[200001], ans[200001];
set<int> d[200001], s;
queue<int> q;
int main() {
scanf("%d%d%d", &n, &m, &k);
for (int i = 1; i <= m; i++)
scanf("%d%d", &x[i], &y[i]), d[x[i]].insert(y[i]), d[y[i]].insert(x[i]);
for (int i = 1; i <= n; i++)
if... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
void decreaseKey(set<pair<int, int> > &pairs, map<int, set<int> > &data,
int decKey, int remVert) {
pair<int, int> removePair;
removePair.first = data[decKey].size();
removePair.second = decKey;
pairs.erase(removePair);
data[decKey].erase(remVert)... |
1037_E. Trips | There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* ... | {
"input": [
"4 4 2\n2 3\n1 2\n1 3\n1 4\n",
"5 8 2\n2 1\n4 2\n5 4\n5 2\n4 3\n5 1\n4 1\n3 2\n",
"5 7 2\n1 5\n3 2\n2 5\n3 4\n1 2\n5 3\n1 3\n"
],
"output": [
"0\n0\n3\n3\n",
"0\n0\n0\n3\n3\n4\n4\n5\n",
"0\n0\n0\n0\n3\n4\n4\n"
]
} | {
"input": [
"16 20 2\n10 3\n5 3\n10 5\n12 7\n7 6\n9 12\n9 6\n1 10\n11 16\n11 1\n16 2\n10 2\n14 4\n15 14\n4 13\n13 15\n1 8\n7 15\n1 7\n8 15\n",
"2 1 1\n2 1\n"
],
"output": [
"0\n0\n3\n3\n3\n3\n7\n7\n7\n7\n7\n11\n11\n11\n11\n15\n15\n15\n15\n16\n",
"2\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 200100;
int n, m, k;
vector<pair<int, int> > veze;
vector<pair<int, int> > graf[MAXN];
void load() {
scanf("%d%d%d", &n, &m, &k);
for (int i = 0; i < m; i++) {
int a, b;
scanf("%d%d", &a, &b);
a--;
b--;
veze.push_back(pair<int, int>(... |
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