description stringlengths 171 4k | code stringlengths 94 3.98k | normalized_code stringlengths 57 4.99k |
|---|---|---|
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | import sys
n, k = map(int, input().split())
M = []
A = []
B = []
for i in range(n):
t, a, b = map(int, input().split())
if a == 1 and b == 1:
M.append(t)
elif a == 1:
A.append(t)
elif b == 1:
B.append(t)
if min(len(M) + len(A), len(M) + len(B)) < k:
print(-1)
sys.exit()
A.append(0)
B.append(0)
M.append(0)
A.sort()
B.sort()
M.sort()
for i in range(len(A) - 1):
A[i + 1] += A[i]
for i in range(len(B) - 1):
B[i + 1] += B[i]
for i in range(len(M) - 1):
M[i + 1] += M[i]
ans = float("inf")
for i in range(k + 1):
j = k - i
if len(A) > i and len(B) > i and len(M) > j:
ans = min(ans, A[i] + B[i] + M[j])
print(ans) | IMPORT ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
time = []
a = []
b = []
count_a = 0
count_b = 0
for i in range(n):
t, a1, b1 = map(int, input().split())
time.append(t)
a.append(a1)
b.append(b1)
if a1 == 1:
count_a += 1
if b1 == 1:
count_b += 1
if count_a < k or count_b < k:
print(-1)
else:
both = []
a_not_b = []
b_not_a = []
for i in range(n):
if a[i] == 1 and b[i] == 1:
both.append(time[i])
elif a[i] == 1 and b[i] == 0:
a_not_b.append(time[i])
elif a[i] == 0 and b[i] == 1:
b_not_a.append(time[i])
a_not_b.sort()
b_not_a.sort()
both_len = len(both)
a_not_b_len = len(a_not_b)
b_not_a_len = len(b_not_a)
final = []
req_len = min(a_not_b_len, b_not_a_len)
for i in range(req_len):
final.append(a_not_b[i] + b_not_a[i])
final.extend(both)
final.sort()
if len(final) < k:
print(-1)
else:
ans = 0
for i in range(k):
ans += final[i]
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | import sys
input = sys.stdin.buffer.readline
def solution():
a = []
b = []
c = []
n, k = map(int, input().split())
for i in range(n):
t, x, y = map(int, input().split())
if x == 1 and y == 1:
c.append(t)
elif x == 1:
a.append(t)
elif y == 1:
b.append(t)
a.sort()
b.sort()
c.sort()
ans = 0
if len(a) + len(c) < k or len(b) + len(c) < k:
print(-1)
else:
z = 0
i = 0
j = 0
while k > 0:
if i < len(c) and j < min(len(a), len(b)):
if a[j] + b[j] <= c[i]:
ans += a[j] + b[j]
j += 1
else:
ans += c[i]
i += 1
elif i < len(c):
ans += c[i]
i += 1
else:
ans += a[j] + b[j]
j += 1
k -= 1
print(ans)
solution() | IMPORT ASSIGN VAR VAR FUNC_DEF ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | from sys import stdin, stdout
input = stdin.readline
print = stdout.write
n, k = map(int, input().split())
alice, bob, together = [], [], []
for _ in range(n):
t, a, b = map(int, input().split())
if a and b:
together += (t,)
elif a:
alice += (t,)
elif b:
bob += (t,)
sa, sb, st = len(alice), len(bob), len(together)
if sa + st < k or sb + st < k:
print("-1")
exit()
sa -= 1
sb -= 1
st -= 1
alice.sort(reverse=True)
bob.sort(reverse=True)
together.sort(reverse=True)
time = 0
for i in range(k):
if sa < 0 or sb < 0:
time += together[st]
st -= 1
elif st < 0:
time += alice[sa] + bob[sb]
sa -= 1
sb -= 1
elif together[st] <= alice[sa] + bob[sb]:
time += together[st]
st -= 1
else:
time += alice[sa] + bob[sb]
sa -= 1
sb -= 1
print(str(time)) | ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR VAR VAR IF VAR VAR VAR IF VAR VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
tab = [list(map(int, input().split())) for _ in range(n)]
x = []
y = []
z = []
for i in range(n):
if tab[i][1] == tab[i][2] == 1:
x.append(tab[i])
elif tab[i][1] == 1:
y.append(tab[i])
elif tab[i][2] == 1:
z.append(tab[i])
x.sort()
y.sort()
z.sort()
sx = [0] * (len(x) + 1)
sy = [0] * (len(y) + 1)
sz = [0] * (len(z) + 1)
for i in range(1, len(x) + 1):
sx[i] = x[i - 1][0]
for i in range(1, len(y) + 1):
sy[i] = y[i - 1][0]
for i in range(1, len(z) + 1):
sz[i] = z[i - 1][0]
for i in range(1, len(sx)):
sx[i] += sx[i - 1]
for i in range(1, len(sy)):
sy[i] += sy[i - 1]
for i in range(1, len(sz)):
sz[i] += sz[i - 1]
ans = 10**18
for i in range(k + 1):
if not (0 <= i <= len(x) and 0 <= k - i <= len(y) and 0 <= k - i <= len(z)):
continue
res = sx[i] + sy[k - i] + sz[k - i]
ans = min(ans, res)
if ans == 10**18:
print(-1)
else:
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF NUMBER VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR BIN_OP NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
al = []
bob = []
both = []
for i in range(n):
a, b, c = map(int, input().split())
if b == 1 and c == 1:
both.append(a)
elif b == 1:
al.append(a)
elif c == 1:
bob.append(a)
tp = len(both)
la = len(al)
ob = len(bob)
a = la + tp
b = ob + tp
if a < k or b < k:
print("-1")
else:
both.sort()
al.sort()
bob.sort()
ans = 0
c = 0
d = 0
for i in range(k):
z = 20001
m = 20001
if c < tp:
z = both[c]
if d < la and d < ob:
m = al[d] + bob[d]
if z < m:
ans += z
c += 1
else:
ans += m
d += 1
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR IF VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR IF VAR VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
both, alice, bob = [], [], []
for _ in range(n):
t, a, b = map(int, input().split())
if a == 1 and b == 1:
both.append(t)
elif a == 1 and b == 0:
alice.append(t)
elif a == 0 and b == 1:
bob.append(t)
both.sort(reverse=True)
alice.sort(reverse=True)
bob.sort(reverse=True)
cnt = time = 0
while cnt < k and both and alice and bob:
if both[-1] <= alice[-1] + bob[-1]:
time += both.pop()
else:
time += alice.pop() + bob.pop()
cnt += 1
while cnt < k and both:
time += both.pop()
cnt += 1
while cnt < k and alice and bob:
time += alice.pop() + bob.pop()
cnt += 1
if cnt >= k:
print(time)
else:
print(-1) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR NUMBER WHILE VAR VAR VAR VAR VAR IF VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER WHILE VAR VAR VAR VAR FUNC_CALL VAR VAR NUMBER WHILE VAR VAR VAR VAR VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().strip().split())
not_possible = False
alice_only, bob_only, both = [], [], []
alice_likes, bob_likes = 0, 0
total_time = 0
for _ in range(n):
t, alice, bob = map(int, input().strip().split())
alice_likes += alice
bob_likes += bob
if alice + bob == 2:
both.append(t)
elif alice:
alice_only.append(t)
elif bob:
bob_only.append(t)
if alice + bob > 0:
total_time += t
if alice_likes < k or bob_likes < k:
not_possible = True
alice_remove = alice_likes - k
bob_remove = bob_likes - k
both_remove = min(alice_remove, bob_remove)
alice_remove -= both_remove
bob_remove -= both_remove
both.sort(reverse=True)
alice_only.sort(reverse=True)
bob_only.sort(reverse=True)
a, b = 0, 0
while a < len(alice_only) and alice_remove > 0:
total_time -= alice_only[a]
a += 1
alice_remove -= 1
while b < len(bob_only) and bob_remove > 0:
total_time -= bob_only[b]
b += 1
bob_remove -= 1
alice_only = alice_only[a:]
bob_only = bob_only[b:]
ab, o = 0, 0
while both_remove > 0:
curr_ab, curr_both = 0, 0
if ab < len(alice_only) and ab < len(bob_only):
curr_ab = alice_only[ab] + bob_only[ab]
if o < len(both):
curr_both = both[o]
if curr_ab + curr_both == 0:
break
if curr_ab > curr_both:
total_time -= curr_ab
ab += 1
else:
total_time -= curr_both
o += 1
both_remove -= 1
alice_remove += both_remove
bob_remove += both_remove
a, b = ab, ab
while a < len(alice_only) and alice_remove > 0:
total_time -= alice_only[a]
a += 1
alice_remove -= 1
while b < len(bob_only) and bob_remove > 0:
total_time -= bob_only[b]
b += 1
bob_remove -= 1
if not_possible:
print(-1)
else:
print(total_time) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR LIST LIST LIST ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR VAR VAR VAR VAR IF BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER VAR VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER WHILE VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR NUMBER NUMBER WHILE VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR IF VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR WHILE VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | t = 1
for _ in range(t):
n, k = map(int, input().split())
arr = []
for i in range(n):
a, b, c = map(int, input().split())
arr.append((a, b, c))
arr = sorted(arr, key=lambda x: x[0])
ans = 0
x = y = k
counter = 0
temp1 = []
temp2 = []
while counter < n:
if x <= 0 and y <= 0:
break
if arr[counter][1] == 0 and arr[counter][2] == 0:
counter += 1
continue
ans += arr[counter][0]
x -= arr[counter][1]
y -= arr[counter][2]
if arr[counter][1] and not arr[counter][2]:
temp1.append(arr[counter][0])
elif not arr[counter][1] and arr[counter][2]:
temp2.append(arr[counter][0])
counter += 1
x_d = min(len(temp1), len(temp2))
ans -= sum(temp1) + sum(temp2)
temp1 = temp1[0:x_d]
temp2 = temp2[0:x_d]
ans += sum(temp1) + sum(temp2)
if len(temp1) > 0 and len(temp2) > 0:
temp1 = sorted(temp1)
temp2 = sorted(temp2)
for i in range(counter, n):
if arr[i][1] == 1 and arr[i][2] == 1:
if arr[i][0] < temp1[-1] + temp2[-1]:
ans += arr[i][0] - (temp1[-1] + temp2[-1])
temp1.pop()
temp2.pop()
if not (len(temp1) and len(temp2)):
break
if x > 0 or y > 0:
print(-1)
else:
print(ans) | ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST WHILE VAR VAR IF VAR NUMBER VAR NUMBER IF VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR NUMBER IF VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER VAR ASSIGN VAR VAR NUMBER VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER IF VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | arr = [int(x) for x in input().split()]
size = arr[0]
total = arr[1]
book_dict = {"2": [], "10": [], "01": []}
for i in range(size):
book_arr = [int(x) for x in input().split()]
if book_arr[1] == 1 and book_arr[2] == 1:
book_dict["2"].append(book_arr[0])
elif book_arr[1] == 1 and book_arr[2] == 0:
book_dict["10"].append(book_arr[0])
elif book_arr[1] == 0 and book_arr[2] == 1:
book_dict["01"].append(book_arr[0])
count10 = len(book_dict["10"])
count01 = len(book_dict["01"])
count2 = len(book_dict["2"])
result = 0
book_dict["10"].sort()
book_dict["01"].sort()
if count10 + count2 >= total and count01 + count2 >= total:
for i in range(min(count01, count10)):
book_dict["2"].append(book_dict["01"][i] + book_dict["10"][i])
book_dict["2"].sort()
result = sum(book_dict["2"][:total])
print(result)
else:
print(-1) | ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR DICT STRING STRING STRING LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR STRING ASSIGN VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING IF BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR STRING BIN_OP VAR STRING VAR VAR STRING VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR STRING VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, m, k = input().split(" ")
n, m, k = int(n), int(m), int(k)
A = []
B = []
C = []
D = []
Ainmax = 0
Binmax = 0
Cinmax = 0
Dinmax = 0
for i in range(n):
entry = input().split(" ")
if entry[1] == "1" and entry[2] == "1":
C.append([int(entry[0]), i + 1])
Cinmax += 1
elif entry[1] == "1" and entry[2] == "0":
A.append([int(entry[0]), i + 1])
Ainmax += 1
elif entry[1] == "0" and entry[2] == "1":
B.append([int(entry[0]), i + 1])
Binmax += 1
else:
D.append([int(entry[0]), i + 1])
Dinmax += 1
A.sort(key=lambda x: x[0])
B.sort(key=lambda x: x[0])
C.sort(key=lambda x: x[0])
D.sort(key=lambda x: x[0])
mi = min(Ainmax, Binmax)
if len(C) + mi < k:
print(-1)
elif len(C) < k and 2 * k - len(C) > m:
print(-1)
else:
time = 0
Ain = 0
Bin = 0
Cin = 0
Din = 0
ABinmax = min(mi, m - k)
for i in range(k):
if Ain == ABinmax:
Cin += 1
elif Cin == Cinmax or A[Ain][0] + B[Bin][0] <= C[Cin][0]:
Ain += 1
Bin += 1
else:
Cin += 1
for i in range(m - Ain - Bin - Cin):
pot = []
if Ain < Ainmax:
pot.append([A[Ain][0], "Ain+=1"])
if Bin < Binmax:
pot.append([B[Bin][0], "Bin+=1"])
if Cin < Cinmax:
pot.append([C[Cin][0], "Cin+=1"])
if Din < Dinmax:
pot.append([D[Din][0], "Din+=1"])
if Ain < Ainmax and Bin < Binmax and Cin != 0:
pot.append([A[Ain][0] + B[Bin][0] - C[Cin - 1][0], "Ain+=1;Bin+=1;Cin-=1"])
minpot = 0
for j in range(len(pot) - 1):
if pot[minpot][0] > pot[j + 1][0]:
minpot = j + 1
exec(pot[minpot][1])
for i in range(Ain):
time += A[i][0]
for i in range(Bin):
time += B[i][0]
for i in range(Cin):
time += C[i][0]
for i in range(Din):
time += D[i][0]
print(time)
for i in range(Ain):
print(A[i][1], end=" ")
for i in range(Bin):
print(B[i][1], end=" ")
for i in range(Cin):
print(C[i][1], end=" ")
for i in range(Din):
print(D[i][1], end=" ") | ASSIGN VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING IF VAR NUMBER STRING VAR NUMBER STRING EXPR FUNC_CALL VAR LIST FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER STRING VAR NUMBER STRING EXPR FUNC_CALL VAR LIST FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER STRING VAR NUMBER STRING EXPR FUNC_CALL VAR LIST FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR LIST FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR VAR VAR BIN_OP BIN_OP NUMBER VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER IF VAR VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR ASSIGN VAR LIST IF VAR VAR EXPR FUNC_CALL VAR LIST VAR VAR NUMBER STRING IF VAR VAR EXPR FUNC_CALL VAR LIST VAR VAR NUMBER STRING IF VAR VAR EXPR FUNC_CALL VAR LIST VAR VAR NUMBER STRING IF VAR VAR EXPR FUNC_CALL VAR LIST VAR VAR NUMBER STRING IF VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR LIST BIN_OP BIN_OP VAR VAR NUMBER VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER STRING FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER STRING FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER STRING FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER STRING |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = [int(k) for k in input().split()]
tog = []
ali = []
bli = []
time = 0
size = 0
for i in range(n):
t, a, b = [int(k) for k in input().split()]
if a & b == 1:
tog.append(t)
else:
if a == 1:
ali.append(t)
if b == 1:
bli.append(t)
asi = len(tog) + len(ali)
bsi = len(tog) + len(bli)
if k > min(asi, bsi):
print(-1)
else:
ali.sort()
bli.sort()
for i in range(min(len(ali), len(bli))):
tog.append(ali[i] + bli[i])
tog.sort()
for i in range(k):
time += tog[i]
print(time) | ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | from sys import stdin
input = stdin.readline
def answer():
if n3 + n1 < k or n3 + n2 < k:
return -1
i, j = 0, 0
ans = 0
for take in range(k):
if i >= n1 or i >= n2:
ans += common[j]
j += 1
elif j >= n3:
ans += a[i] + b[i]
i += 1
elif a[i] + b[i] > common[j]:
ans += common[j]
j += 1
else:
ans += a[i] + b[i]
i += 1
return ans
n, k = map(int, input().split())
a, b, common = [], [], []
for i in range(n):
t, x, y = map(int, input().split())
if x and y:
common.append(t)
elif x == 1 and y == 0:
a.append(t)
elif x == 0 and y == 1:
b.append(t)
common.sort()
a.sort()
b.sort()
n1, n2, n3 = len(a), len(b), len(common)
print(answer()) | ASSIGN VAR VAR FUNC_DEF IF BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR RETURN NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER IF BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER RETURN VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = tuple(map(int, input().split()))
alice = []
bob = []
both = []
for _ in range(n):
t, a, b = tuple(map(int, input().split()))
if a == 1 and b == 1:
both.append(t)
elif a == 1:
alice.append(t)
elif b == 1:
bob.append(t)
both.sort()
alice.sort()
bob.sort()
remain = 0
if k <= len(both):
result = sum(both[:k])
else:
remain = k - len(both)
if remain > len(bob) or remain > len(alice):
result = -1
else:
result = sum(both)
result += sum(alice[:remain]) + sum(bob[:remain])
index = max(0, k - len(both))
lenbob = len(bob)
lenalice = len(alice)
if result != -1:
for i in range(min(k, len(both)) - 1, -1, -1):
if index > lenbob - 1 or index > lenalice - 1:
break
newresult = result - both[i] + alice[index] + bob[index]
index += 1
if newresult < result:
result = newresult
else:
break
print(result) | ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | from sys import exit, stdin
input = stdin.readline
def i():
return input()
def ii():
return int(input())
def iis():
return map(int, input().split())
def liis():
return list(map(int, input().split()))
def print_array(a):
print(" ".join(map(str, a)))
n, k = iis()
alice = []
bob = []
both = []
for _ in range(n):
t, a, b = iis()
if a == 1 and b == 0:
alice.append(t)
elif a == 0 and b == 1:
bob.append(t)
elif a == 1 and b == 1:
both.append(t)
alice = sorted(alice)[::-1]
bob = sorted(bob)[::-1]
both = sorted(both)[::-1]
time = 0
livros = 0
while k > 0 and (len(both) != 0 or len(alice) != 0 and len(bob) != 0):
if len(both) and len(alice) and len(bob):
separado = alice[-1] + bob[-1]
junto = both[-1]
if junto < separado:
time += junto
both.pop()
else:
time += separado
alice.pop()
bob.pop()
elif len(both):
time += both.pop()
elif len(alice) and len(bob):
time += alice.pop()
time += bob.pop()
else:
continue
k -= 1
if k == 0:
print(time)
else:
print(-1) | ASSIGN VAR VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | import sys
input = sys.stdin.readline
inp, ip = lambda: int(input()), lambda: [int(w) for w in input().split()]
n, k = ip()
t, a, b = [0] * n, [0] * n, [0] * n
both = []
alice, bob = [], []
for i in range(n):
t[i], a[i], b[i] = ip()
if a[i] and b[i]:
both.append(t[i])
elif a[i]:
alice.append(t[i])
elif b[i]:
bob.append(t[i])
if a.count(1) < k or b.count(1) < k:
print(-1)
exit()
alice.sort()
bob.sort()
m = min(len(alice), len(bob))
for i in range(m):
both.append(alice[i] + bob[i])
both.sort()
print(sum(both[:k])) | IMPORT ASSIGN VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR VAR VAR BIN_OP LIST NUMBER VAR BIN_OP LIST NUMBER VAR BIN_OP LIST NUMBER VAR ASSIGN VAR LIST ASSIGN VAR VAR LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | import sys
input = sys.stdin.readline
n, minn = map(int, input().split())
a = []
ca, cb = 0, 0
for i in range(n):
t, x, y = map(int, input().split())
if x == 1:
ca += 1
if y == 1:
cb += 1
a.append((t, x, y))
if ca < minn or cb < minn:
print(-1)
return
ca, cb = 0, 0
alice = []
bob = []
both = []
for i in range(n):
if a[i][1] == a[i][2] == 1:
both.append(a[i][0])
elif a[i][1] != a[i][2]:
if a[i][1] == 1:
alice.append(a[i][0])
else:
bob.append(a[i][0])
alice.sort()
bob.sort()
both.sort()
i, j, k = 0, 0, 0
ans = 0
while ca < minn or cb < minn:
if (i < len(alice) and j < len(bob) and k < len(both)) and alice[i] + bob[
j
] <= both[k]:
ans += alice[i] + bob[j]
ca += 1
cb += 1
i += 1
j += 1
elif (i < len(alice) and j < len(bob) and k < len(both)) and alice[i] + bob[
j
] >= both[k]:
ans += both[k]
ca += 1
cb += 1
k += 1
else:
break
if i == len(alice):
while ca < minn:
ans += both[k]
k += 1
ca += 1
cb += 1
if j == len(bob):
while cb < minn:
ans += both[k]
k += 1
ca += 1
cb += 1
if k == len(both):
while ca < minn:
ans += alice[i]
i += 1
ca += 1
while cb < minn:
ans += bob[j]
j += 1
cb += 1
print(ans) | IMPORT ASSIGN VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR NUMBER VAR VAR NUMBER IF VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR WHILE VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR WHILE VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR WHILE VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER WHILE VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = list(map(int, input().split()))
a = []
b = []
comb = []
for _ in range(n):
arr = list(map(int, input().split()))
if arr[1] == 1 and arr[2] == 1:
comb.append(arr[0])
elif arr[1] == 1:
a.append(arr[0])
elif arr[2] == 1:
b.append(arr[0])
if len(a) + len(comb) < k or len(b) + len(comb) < k:
print(-1)
exit()
a.sort()
b.sort()
comb.sort()
ans = 0
cnt = 0
i = 0
j = 0
sep = min(len(a), len(b))
joint = len(comb)
while cnt < k:
if i >= sep:
ans += comb[j]
j += 1
cnt += 1
elif j >= joint:
ans += a[i] + b[i]
i += 1
cnt += 1
elif a[i] + b[i] <= comb[j]:
ans += a[i] + b[i]
i += 1
cnt += 1
else:
ans += comb[j]
j += 1
cnt += 1
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR IF VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF BIN_OP VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
d = {"a": [], "b": [], "c": []}
for _ in range(n):
t, a, b = map(int, input().split())
if (a, b) == (1, 0):
d["a"].append(t)
if (a, b) == (0, 1):
d["b"].append(t)
if (a, b) == (1, 1):
d["c"].append(t)
d["a"].sort()
d["b"].sort()
d["c"].sort()
a = d["a"]
b = d["b"]
c = d["c"]
a1 = [0]
b1 = [0]
c1 = [0]
num = 0
for el in a:
num += el
a1.append(num)
num = 0
for el in b:
num += el
b1.append(num)
num = 0
for el in c:
num += el
c1.append(num)
if len(a) + len(c) < k or len(b) + len(c) < k:
print(-1)
else:
times = []
i = 0
while i <= k:
try:
temp = c1[i] + a1[k - i] + b1[k - i]
times.append(temp)
except IndexError:
pass
finally:
i += 1
print(min(times)) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR DICT STRING STRING STRING LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING VAR IF VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING VAR IF VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING ASSIGN VAR VAR STRING ASSIGN VAR VAR STRING ASSIGN VAR VAR STRING ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
A, B, C = [], [], []
for i in range(n):
t, x, y = map(int, input().split())
if x == 1 and y == 1:
C.append(t)
continue
if x == 1:
A.append(t)
continue
if y == 1:
B.append(t)
def fun(l):
pre = [0]
for i in l:
pre.append(pre[-1] + i)
return pre
A.sort()
B.sort()
C.sort()
a = fun(A)
b = fun(B)
c = fun(C)
if len(A) + len(C) < k or len(B) + len(C) < k:
print(-1)
else:
ans = 999999999999999999999999999
for i in range(min(len(C), k) + 1):
if k - i < len(a) and k - i < len(b):
ans = min(ans, c[i] + a[k - i] + b[k - i])
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR LIST NUMBER FOR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR RETURN VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR VAR NUMBER IF BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = [int(x) for x in input().split()]
at = []
bt = []
ct = []
for i in range(n):
ti, ai, bi = [int(x) for x in input().split()]
if ai == 1 and bi == 1:
ct.append(ti)
elif ai == 1:
at.append(ti)
elif bi == 1:
bt.append(ti)
else:
continue
if len(ct) + len(at) < k or len(ct) + len(bt) < k:
print(-1)
else:
ci = 0
ai = 0
al = k
ans = 0
ct.sort()
at.sort()
bt.sort()
i = 0
while i < len(ct) and al > 0 and ai < min(len(at), len(bt)):
if ct[i] < at[ai] + bt[ai]:
ans += ct[i]
i += 1
else:
ans += at[ai] + bt[ai]
ai += 1
al -= 1
if i == len(ct) and al != 0:
bl = al
ans += sum(at[ai : ai + al]) + sum(bt[ai : ai + al])
elif ai == min(len(at), len(bt)) and al > 0:
ans += sum(ct[i : i + al])
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR IF VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | import sys
input = sys.stdin.readline
f = lambda: list(map(int, input().strip("\n").split()))
n, k = f()
_11 = []
_01 = []
_10 = []
for _ in range(n):
t, a, b = f()
if a and b:
_11.append(t)
elif a:
_10.append(t)
elif b:
_01.append(t)
_01.sort()
_10.sort()
_11.sort()
for i in range(1, len(_01)):
_01[i] += _01[i - 1]
for i in range(1, len(_10)):
_10[i] += _10[i - 1]
for i in range(1, len(_11)):
_11[i] += _11[i - 1]
ans = 3 * 1000000000.0
if len(_01) >= k and len(_10) >= k:
ans = min(ans, _01[k - 1] + _10[k - 1])
for i in range(len(_11)):
if i + 1 < k and len(_01) >= k - i - 1 and len(_10) >= k - i - 1:
ans = min(ans, _11[i] + _01[k - i - 2] + _10[k - i - 2])
elif len(_11) >= k:
ans = min(ans, _11[k - 1])
break
print(-1 if ans == 3 * 1000000000.0 else ans) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP NUMBER NUMBER NUMBER VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
la, lb, lc = [], [], []
for _ in range(n):
t, a, b = map(int, input().split())
if a == b == 1:
lc.append(t)
elif a == 1 and b == 0:
la.append(t)
elif a == 0 and b == 1:
lb.append(t)
la.sort()
lb.sort()
lc.sort()
t = 0
ln = len(lc)
if ln >= k:
t += sum(lc[:k])
a, b = k, k
ln = k
else:
t += sum(lc)
a, b = ln, ln
if a + len(la) < k or b + len(lb) < k:
print(-1)
else:
t += sum(la[: k - a]) + sum(lb[: k - b])
u, v = k - a, k - b
while u < len(la) and v < len(lb) and ln > 0:
if t - lc[ln - 1] + la[u] + lb[v] < t:
t = t - lc[ln - 1] + la[u] + lb[v]
u += 1
v += 1
ln -= 1
else:
break
print(t) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR IF BIN_OP VAR FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER IF BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, m, k = map(int, input().split())
ab = []
a = []
b = []
other = []
l = [list(map(int, input().split())) for i in range(n)]
for i in range(n):
t, c, d = l[i]
if c and d == 0:
a.append([t, i + 1])
elif d and c == 0:
b.append([t, i + 1])
elif c * d:
ab.append([t, i + 1])
else:
other.append([t, i + 1])
a.sort()
b.sort()
ab.sort()
other.sort()
la = len(a)
lb = len(b)
lab = len(ab)
lo = len(other)
a.append([float("INF"), -1])
b.append([float("INF"), -1])
ab.append([float("INF"), -1])
other.append([float("INF"), -1])
if la < lb:
la, lb = lb, la
a, b = b, a
ans = float("INF")
count = 0
ia = 0
ib = 0
iab = 0
io = 0
ana = 0
anb = 0
anab = 0
ano = 0
for i in range(lab + 1):
if k - i > lb:
continue
if 2 * k - i > m:
continue
if i + la + lb + lo < m:
continue
if ia > 0:
ia -= 1
count -= a[ia][0]
if ib > 0:
ib -= 1
count -= b[ib][0]
if io > 0:
io -= 1
count -= other[io][0]
while ia < la and ia < k - i:
count += a[ia][0]
ia += 1
while ib < lb and ib < k - i:
count += b[ib][0]
ib += 1
while iab < lab and iab < i:
count += ab[iab][0]
iab += 1
while ia + ib + iab + io < m:
na = a[ia][0]
nb = b[ib][0]
no = other[io][0]
mi = min(na, nb, no)
if mi == na:
count += na
ia += 1
elif mi == nb:
count += nb
ib += 1
else:
count += no
io += 1
if count < ans:
ans = count
ana = ia
anb = ib
anab = iab
ano = io
if ans == float("INF"):
print(-1)
else:
print(ans)
l = []
for i in range(ana):
l.append(a[i][1])
for i in range(anb):
l.append(b[i][1])
for i in range(anab):
l.append(ab[i][1])
for i in range(ano):
l.append(other[i][1])
print(*l) | ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR LIST VAR BIN_OP VAR NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR LIST VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR EXPR FUNC_CALL VAR LIST VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR LIST VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR LIST FUNC_CALL VAR STRING NUMBER EXPR FUNC_CALL VAR LIST FUNC_CALL VAR STRING NUMBER EXPR FUNC_CALL VAR LIST FUNC_CALL VAR STRING NUMBER EXPR FUNC_CALL VAR LIST FUNC_CALL VAR STRING NUMBER IF VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR VAR IF BIN_OP BIN_OP NUMBER VAR VAR VAR IF BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER VAR VAR VAR NUMBER WHILE VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER WHILE VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER WHILE VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER WHILE BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | def solve():
global B, K
both = [b[0] for b in B if b[1] and b[2]]
alice = sorted([b[0] for b in B if b[1] and not b[2]])
bob = sorted([b[0] for b in B if b[2] and not b[1]])
for i in range(min(len(alice), len(bob))):
both.append(alice[i] + bob[i])
if len(both) < K:
return -1
return sum(sorted(both)[:K])
N, K = map(int, input().split())
B = [tuple(map(int, input().split())) for _ in range(N)]
sol = solve()
print(sol) | FUNC_DEF ASSIGN VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR RETURN NUMBER RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
alice = []
bob = []
both = []
for _ in range(n):
t, a, b = map(int, input().split())
if a == 1 and b == 1:
both.append((t, a, b))
elif a == 1 and b == 0:
alice.append((t, a, b))
elif a == 0 and b == 1:
bob.append((t, a, b))
alice.sort()
bob.sort()
both.sort()
ans = 0
ca = 0
cb = 0
ia = 0
ib = 0
iboth = 0
while ca < k and cb < k and (ia < len(alice) and ib < len(bob) or iboth < len(both)):
if (
iboth >= len(both)
or ia < len(alice)
and ib < len(bob)
and alice[ia][0] + bob[ib][0] < both[iboth][0]
):
ans += alice[ia][0] + bob[ib][0]
ia += 1
ib += 1
else:
ans += both[iboth][0]
iboth += 1
ca += 1
cb += 1
if cb >= k and ca >= k:
print(ans)
else:
print(-1) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
a, b, ab = [], [], []
for _ in range(n):
t, x, y = map(int, input().split())
if x == 1 and y == 1:
ab.append(t)
elif x == 1 and y == 0:
a.append(t)
elif x == 0 and y == 1:
b.append(t)
if len(ab) + min(len(a), len(b)) < k:
print(-1)
else:
cost, ind, tog = 0, 0, 0
a.sort()
b.sort()
ab.sort()
for i in range(k):
if tog == len(ab):
cost += sum(a[ind : ind + (k - i)]) + sum(b[ind : ind + (k - i)])
break
elif ind == min(len(a), len(b)):
cost += sum(ab[tog : tog + (k - i)])
break
elif a[ind] + b[ind] < ab[tog]:
cost += a[ind] + b[ind]
ind += 1
else:
cost += ab[tog]
tog += 1
print(cost) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR VAR IF VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR VAR IF BIN_OP VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
alice = []
bob = []
both = []
for _ in range(n):
t, a, b = map(int, input().split())
if a and b:
both.append(t)
elif a:
alice.append(t)
elif b:
bob.append(t)
if len(alice) + len(both) < k or len(bob) + len(both) < k:
print(-1)
else:
alice.sort()
bob.sort()
both.sort()
i = j = 0
ans = 0
while i < len(alice) and i < len(bob) and j < len(both) and k > 0:
if alice[i] + bob[i] < both[j]:
ans += alice[i] + bob[i]
i += 1
else:
ans += both[j]
j += 1
k -= 1
if k:
if j >= len(both):
while k > 0:
ans += alice[i] + bob[i]
i += 1
k -= 1
else:
while k > 0:
ans += both[j]
j += 1
k -= 1
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER IF BIN_OP VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR IF VAR FUNC_CALL VAR VAR WHILE VAR NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k1 = map(int, input().split())
c = []
a = []
b = []
for i in range(n):
x, y, z = map(int, input().split())
if y == z == 1:
c.append(x)
elif y == 1 and z == 0:
a.append(x)
elif y == 0 and z == 1:
b.append(x)
a.sort()
b.sort()
c.sort()
ka = k1
kb = k1
i, j, k = 0, 0, 0
ans = 0
while ka > 0 and kb > 0 and i < len(a) and j < len(b) and k < len(c):
if a[i] + b[j] <= c[k]:
ans += a[i] + b[j]
ka -= 1
kb -= 1
i += 1
j += 1
else:
ans += c[k]
ka -= 1
kb -= 1
k += 1
if i >= len(a) and ka > 0:
while k < len(c) and ka > 0:
ans += c[k]
k += 1
ka -= 1
kb = max(0, kb - 1)
elif j >= len(b) and kb > 0:
while k < len(c) and kb > 0:
ans += c[k]
k += 1
ka = max(0, ka - 1)
kb -= 1
while i < len(a) and ka > 0:
ans += a[i]
ka -= 1
i += 1
while j < len(b) and kb > 0:
ans += b[j]
kb -= 1
j += 1
if ka == kb == 0:
print(ans)
else:
print(-1) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
z11, z01, z10 = [], [], []
for i in range(n):
t, a, b = map(int, input().split())
if a == 1 and b == 1:
z11.append(t)
elif a == 1:
z10.append(t)
elif b == 1:
z01.append(t)
i, j = min(k, len(z11)), min(k, len(z01), len(z10))
z11.sort()
z10.sort()
z01.sort()
if i + j < k:
print(-1)
exit()
while i + j > k:
if z11[i - 1] > z10[j - 1] + z01[j - 1]:
i -= 1
else:
j -= 1
print(sum(z11[:i]) + sum(z10[:j]) + sum(z01[:j])) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR WHILE BIN_OP VAR VAR VAR IF VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
a, b, t = [], [], []
for i in range(0, n):
tm, c, d = map(int, input().split())
if c == 1 and d == 1:
t.append(tm)
elif c == 1 and d == 0:
a.append(tm)
elif c == 0 and d == 1:
b.append(tm)
if len(t) + len(a) < k or len(t) + len(b) < k:
print(-1)
else:
t.sort()
a.sort()
b.sort()
i = 0
j = 0
ans = 0
fga = 0
fgt = 0
while k != 0:
if len(a) <= j or len(b) <= j:
fga = 1
break
elif len(t) <= i:
fgt = 1
break
elif t[i] > a[j] + b[j]:
ans += a[j] + b[j]
j += 1
elif t[i] <= a[j] + b[j]:
ans += t[i]
i += 1
k -= 1
if fga == 1 and fgt == 0:
while k != 0:
ans += t[i]
i += 1
k -= 1
elif fga == 0 and fgt != 0:
while k != 0:
ans += a[j] + b[j]
j += 1
k -= 1
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR BIN_OP VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER IF VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER WHILE VAR NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
ans = []
c1 = 0
c2 = 0
for i in range(n):
z = list(map(int, input().split()))
if z[1] == 1:
c1 += 1
if z[2] == 1:
c2 += 1
ans.append(z)
if min(c1, c2) < k:
print(-1)
else:
a0 = []
a1 = []
a2 = []
for i in range(len(ans)):
if ans[i][1] == 1 and ans[i][2] == 1:
a0.append(ans[i])
elif ans[i][1] == 1 and ans[i][2] == 0:
a1.append(ans[i])
elif ans[i][1] == 0 and ans[i][2] == 1:
a2.append(ans[i])
p0 = 0
p1 = 0
p2 = 0
c1 = 0
c2 = 0
a0.sort()
a1.sort()
a2.sort()
total = 0
while c1 < k and c2 < k:
if (
p0 < len(a0)
and p1 < len(a1)
and p2 < len(a2)
and a0[p0][0] <= a1[p1][0] + a2[p2][0]
):
total += a0[p0][0]
c1 += 1
c2 += 1
p0 += 1
else:
if p1 < len(a1) and p2 < len(a2):
total += a1[p1][0] + a2[p2][0]
c1 += 1
c2 += 1
p1 += 1
p2 += 1
continue
if p1 == len(a1) or p2 == len(a2):
total += a0[p0][0]
c1 += 1
c2 += 1
p0 += 1
print(total) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER NUMBER VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | import itertools
def good(combo, k):
alice = 0
bob = 0
for t, a, b in combo:
if a == 1:
alice += 1
if b == 1:
bob += 1
return alice >= k and bob >= k
def brute(arr, k):
result = float("inf")
for comboSize in range(len(arr) - 1, 0, -1):
for combo in itertools.combinations(arr, comboSize):
if good(combo, k):
curr = sum(t for t, _, _ in combo)
if curr < result:
result = curr
if result == float("inf"):
return -1
return result
def fast(arr, k):
anums = []
bnums = []
cnums = []
for t, a, b in arr:
if a == 1 and b == 0:
anums.append(t)
elif a == 0 and b == 1:
bnums.append(t)
elif a == 1 and b == 1:
cnums.append(t)
anums.sort()
bnums.sort()
cnums.sort()
result = 0
a, b, c = 0, 0, 0
alice = 0
bob = 0
while alice < k and bob < k:
if a < len(anums) and b < len(bnums):
up1 = anums[a] + bnums[b]
else:
up1 = float("inf")
if c < len(cnums):
up2 = cnums[c]
else:
up2 = float("inf")
if up1 == float("inf") and up2 == float("inf"):
return -1
if up1 < up2:
a += 1
b += 1
result += up1
else:
c += 1
result += up2
alice += 1
bob += 1
return result
def main():
n, k = [int(x) for x in input().split()]
arr = []
for _ in range(n):
t, a, b = [int(x) for x in input().split()]
arr.append((t, a, b))
result = fast(arr, k)
print(result)
main() | IMPORT FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER RETURN VAR VAR VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR FUNC_CALL VAR STRING RETURN NUMBER RETURN VAR FUNC_DEF ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR STRING IF VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR STRING IF VAR FUNC_CALL VAR STRING VAR FUNC_CALL VAR STRING RETURN NUMBER IF VAR VAR VAR NUMBER VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | import sys
input = sys.stdin.readline
def main():
n, k = map(int, input().split())
a_s = []
b_s = []
ab_s = []
for _ in range(n):
t, a, b = map(int, input().split())
if a == b == 1:
ab_s.append(t)
elif a == 1:
a_s.append(t)
elif b == 1:
b_s.append(t)
la = len(a_s)
lb = len(b_s)
lab = len(ab_s)
if la + lab < k or lb + lab < k:
print(-1)
return
a_s.sort()
b_s.sort()
for i in range(min(la, lb)):
tmp = a_s[i] + b_s[i]
ab_s.append(tmp)
ab_s.sort()
print(sum(ab_s[:k]))
main() | IMPORT ASSIGN VAR VAR FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | from sys import stdin
inp = lambda: stdin.readline().strip()
n, k = [int(x) for x in inp().split()]
b = []
for _ in range(n):
b.append([int(x) for x in inp().split()])
both = []
alice = []
bob = []
for i in b:
if i[1] == 1 and i[2] == 1:
both.append(i[0])
elif i[1] == 1:
alice.append(i[0])
elif i[2] == 1:
bob.append(i[0])
if len(alice) + len(both) < k or len(bob) + len(both) < k:
print(-1)
exit()
both.sort()
alice.sort()
bob.sort()
b = 0
ind = 0
cost = 0
liked = 0
minimum = min(len(alice), len(bob))
x = max(k - minimum, 0)
for i in range(x):
if liked == k:
print(cost)
exit()
cost += both[i]
b += 1
liked += 1
while True:
if liked == k:
print(cost)
break
if (
b < len(both)
and ind < len(alice)
and ind < len(bob)
and both[b] <= alice[ind] + bob[ind]
):
cost += both[b]
b += 1
else:
cost += alice[ind] + bob[ind]
ind += 1
liked += 1 | ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER WHILE NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | import sys
input = sys.stdin.readline
n, k = map(int, input().split())
books = [tuple(map(int, input().split())) for _ in range(n)]
books.sort(key=lambda x: (x[0], -x[1], -x[2]))
ans = 0
comm = []
cnt_a = []
cnt_b = []
for t, a, b in books:
if a == 0 and b == 0:
pass
elif a == 1 and b == 1:
if len(cnt_a) + len(comm) < k or len(cnt_b) + len(comm) < k:
comm.append(t)
elif cnt_a and cnt_b and cnt_a[-1] + cnt_b[-1] > t:
cnt_a.pop()
cnt_b.pop()
comm.append(t)
elif a == 1 and len(cnt_a) + len(comm) < k:
cnt_a.append(t)
elif b == 1 and len(cnt_b) + len(comm) < k:
cnt_b.append(t)
else:
pass
while cnt_a and len(cnt_a) + len(comm) > k:
cnt_a.pop()
while cnt_b and len(cnt_b) + len(comm) > k:
cnt_b.pop()
if len(cnt_a) + len(comm) < k or len(cnt_b) + len(comm) < k:
print(-1)
else:
print(sum(cnt_a) + sum(cnt_b) + sum(comm)) | IMPORT ASSIGN VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR IF VAR NUMBER BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR NUMBER BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR WHILE VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR WHILE VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
common_books = []
alice_books = []
bob_books = []
for i in range(n):
duration, alice, bob = map(int, input().split())
if alice and bob:
collection = common_books
elif alice:
collection = alice_books
elif bob:
collection = bob_books
else:
collection = None
if collection is not None:
collection.append(duration)
def get0(collection, idx):
if idx < len(collection):
return collection[idx]
return 0
if len(bob_books) + len(common_books) < k or len(alice_books) + len(common_books) < k:
print(-1)
else:
for collection in (common_books, alice_books, bob_books):
collection.sort()
icom = 0
iali = 0
ibob = 0
ncom = len(common_books)
nali = len(alice_books)
nbob = len(bob_books)
total = 0
for i in range(k):
if (
icom < ncom
and get0(alice_books, iali) + get0(bob_books, ibob) >= common_books[icom]
):
total += common_books[icom]
icom += 1
elif iali < nali and ibob < nbob:
total += alice_books[iali]
total += bob_books[ibob]
iali += 1
ibob += 1
else:
total += common_books[icom]
icom += 1
print(total) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR ASSIGN VAR VAR IF VAR ASSIGN VAR VAR IF VAR ASSIGN VAR VAR ASSIGN VAR NONE IF VAR NONE EXPR FUNC_CALL VAR VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR RETURN VAR VAR RETURN NUMBER IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER FOR VAR VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | book_number, like_num = map(int, input().split())
both_like = []
only_alice = []
only_bob = []
for _ in range(book_number):
t, a, b = map(int, input().split())
if a == b == 1:
both_like.append(t)
elif a == 1:
only_alice.append(t)
elif b == 1:
only_bob.append(t)
i, j = 0, 0
time = 0
current_like = 0
both_like.sort()
only_alice.sort()
only_bob.sort()
while current_like < like_num:
if i < len(both_like) and j < min(len(only_alice), len(only_bob)):
if both_like[i] < only_alice[j] + only_bob[j]:
time += both_like[i]
i += 1
else:
time += only_alice[j] + only_bob[j]
j += 1
current_like += 1
elif i < len(both_like):
time += both_like[i]
i += 1
current_like += 1
elif j < min(len(only_alice), len(only_bob)):
time += only_alice[j] + only_bob[j]
j += 1
current_like += 1
else:
break
if current_like == like_num:
print(time)
else:
print(-1) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR WHILE VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, kk = map(int, input().split())
alice = []
bob = []
both = []
for i in range(n):
t, a, b = map(int, input().split())
if a == 1 and b == 1:
both.append(t)
elif a == 1:
alice.append(t)
elif b == 1:
bob.append(t)
alice.sort()
bob.sort()
both.sort()
count = 0
l = min(len(alice), len(bob))
i = 0
j = 0
k = 0
ans = 0
if len(both) + len(bob) < kk or len(alice) + len(both) < kk:
print(-1)
else:
while j < l and count < kk:
if k >= len(both) or alice[j] + bob[j] < both[k]:
ans += alice[j] + bob[j]
j += 1
count += 1
else:
ans += both[k]
k += 1
count += 1
print(ans + sum(both[k : k + kk - count])) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER WHILE VAR VAR VAR VAR IF VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | def next():
return [int(x) for x in input().split()]
class book:
def __init__(self, t, a, b):
self.t = t
self.a = a
self.b = b
n, k = next()
books = []
for _ in range(n):
t, a, b = next()
books.append(book(t, a, b))
oo = list(map(lambda x: x.t, filter(lambda x: x.a == 1 and x.b == 1, books)))
zo = list(sorted(map(lambda x: x.t, filter(lambda x: x.a == 0 and x.b == 1, books))))
oz = list(sorted(map(lambda x: x.t, filter(lambda x: x.a == 1 and x.b == 0, books))))
mi = min(len(zo), len(oz))
for i in range(mi):
oo.append(zo[i] + oz[i])
if len(oo) < k:
print(-1)
else:
print(sum(sorted(oo)[:k])) | FUNC_DEF RETURN FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR CLASS_DEF FUNC_DEF ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | def func():
n, k = map(int, input().split())
sa, sb, both = [], [], []
for _ in range(n):
t, aLike, bLike = map(int, input().split())
if aLike and bLike:
both.append(t)
elif aLike:
sa.append(t)
elif bLike:
sb.append(t)
sa.sort()
sb.sort()
both.sort()
if min(len(sa), len(sb)) + len(both) < k:
print(-1)
return
ans = 2**32
common = 0 if len(sa) >= k and len(sb) >= k else k - min(len(sa), len(sb))
cur = sum(both[:common]) + sum(sa[: k - common]) + sum(sb[: k - common])
ans = min(cur, ans)
while common < min(k, len(both)):
cur = cur + both[common] - sa[k - common - 1] - sb[k - common - 1]
ans = min(cur, ans)
common += 1
print(ans)
func() | FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | from sys import stdin
n, k = list(map(int, stdin.readline().split()))
alice = []
bob = []
both = []
for _ in range(n):
t, a, b = list(map(int, stdin.readline().split()))
if a == b == 1:
both.append(t)
elif a == 1:
alice.append(t)
elif b == 1:
bob.append(t)
alice = sorted(alice)
bob = sorted(bob)
both = sorted(both)
a, b, c = 0, 0, 0
res = 0
for _ in range(k):
if c >= len(both) and (a >= len(alice) or b >= len(bob)):
res = -1
break
if c >= len(both):
res += alice[a]
res += bob[b]
a += 1
b += 1
elif a >= len(alice) or b >= len(bob):
res += both[c]
c += 1
elif alice[a] + bob[b] < both[c]:
res += alice[a]
res += bob[b]
a += 1
b += 1
elif both[c] <= alice[a] + bob[b]:
res += both[c]
c += 1
print(res) | ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER IF BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
data = []
a_num = 0
b_num = 0
a_array = []
b_array = []
ab_array = []
time_sum = 0
for i in range(n):
time, a, b = map(int, input().split())
if a and b:
a_num += 1
b_num += 1
ab_array.append(time)
elif a:
a_num += 1
a_array.append(time)
elif b:
b_num += 1
b_array.append(time)
if a_num < k or b_num < k:
print(-1)
else:
a_array.sort()
b_array.sort()
while a_array and b_array:
ab_array.append(a_array.pop(0) + b_array.pop(0))
ab_array.sort()
for i in range(k):
time_sum += ab_array[i]
print(time_sum) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR WHILE VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR NUMBER FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | def reading(numb, mass, c1, c2):
if c1 < numb or c2 < numb:
return -1
true_mass1 = []
true_mass2 = []
true_mass3 = []
for i in mass:
if i[1] != 0 or i[2] != 0:
if i[1] == 0:
true_mass1.append(i)
elif i[2] == 0:
true_mass2.append(i)
else:
true_mass3.append(i)
true_mass1.sort(key=lambda p: p[0])
true_mass2.sort(key=lambda p: p[0])
true_mass3.sort(key=lambda p: p[0])
i = 0
j = 0
k = 0
deltai = len(true_mass3)
deltaj = len(true_mass2)
deltak = len(true_mass1)
result = 0
for d in range(numb):
if deltai != i and deltaj != j and deltak != k:
if true_mass3[i][0] <= true_mass2[j][0] + true_mass1[k][0]:
result += true_mass3[i][0]
i += 1
else:
result += true_mass2[j][0] + true_mass1[k][0]
k += 1
j += 1
elif deltai == i:
result += true_mass2[j][0] + true_mass1[k][0]
k += 1
j += 1
elif deltaj == j or deltak == k:
result += true_mass3[i][0]
i += 1
return result
t = [int(i) for i in input().strip().split()]
n = t[1]
count1 = 0
count2 = 0
mass = []
for i in range(t[0]):
string = [int(j) for j in input().strip().split()]
mass.append(string)
if string[1] == 1:
count1 += 1
if string[2] == 1:
count2 += 1
print(reading(n, mass, count1, count2)) | FUNC_DEF IF VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR IF VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR IF VAR NUMBER NUMBER VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
lst = []
for i in range(n):
lst.append(list(map(int, input().split())))
alice = []
bob = []
same = []
for i in lst:
if i[1] == 1 and i[2] == 0:
alice.append(i[0])
elif i[2] == 1 and i[1] == 0:
bob.append(i[0])
elif i[2] == 1 and i[1] == 1:
same.append(i[0])
a = len(alice)
b = len(bob)
s = len(same)
if a + s < k or b + s < k:
print(-1)
else:
alice.sort()
bob.sort()
same.sort()
l = 0
n = 0
ans = 0
count = 0
temp = min(a, b)
while count < k and n < s and l < temp:
if alice[l] + bob[l] < same[n] or n >= s:
ans += alice[l] + bob[l]
l += 1
count += 1
else:
ans += same[n]
count += 1
n += 1
if count != k and n == s:
ans += sum(alice[l : l + k - count]) + sum(bob[l : l + k - count])
else:
ans += sum(same[n : n + k - count])
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR WHILE VAR VAR VAR VAR VAR VAR IF BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
books = []
alice = set()
bob = set()
for i in range(n):
books.append([int(i) for i in input().split()])
if books[-1][1] == 1:
alice.add(i)
if books[-1][2] == 1:
bob.add(i)
if len(alice) < k or len(bob) < k:
print(-1)
else:
time, a, b = 0, 0, 0
temp = alice | bob
diff = temp.difference(alice & bob)
temp_alice = []
temp_bob = []
for i in diff:
if i in alice:
temp_alice += [books[i]]
else:
temp_bob += [books[i]]
temp_alice.sort()
temp_bob.sort()
temp_mass = []
for i in range(min(len(temp_bob), len(temp_alice))):
temp_mass.append([temp_alice[i][0] + temp_bob[i][0], 1, 1])
temp_mass = temp_mass + [books[i] for i in temp if i not in diff]
temp_mass.sort(key=lambda x: (x[1] + x[2], -x[0]), reverse=True)
for i in temp_mass:
time += i[0]
a += i[1]
b += i[2]
if a >= k and b >= k:
break
print(time) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR IF VAR VAR VAR LIST VAR VAR VAR LIST VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR LIST BIN_OP VAR VAR NUMBER VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER NUMBER FOR VAR VAR VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | import sys
n, k = [int(e) for e in input().split(" ")]
books = list()
books_map = dict()
for i in range(n):
t, a, b = [int(e) for e in input().split(" ")]
books.append((t, a, b, i))
books_map[i] = t, a, b
books = sorted(books, key=lambda x: x[0] * (1 if x[1] == 1 else 10000000.0))
alice_pay_time = 0
alice_liked_books = list()
for book in books:
if len(alice_liked_books) >= k:
break
if book[1] == 1 and book[2] == 0:
alice_pay_time += book[0]
alice_liked_books.append(book[3])
books = sorted(books, key=lambda x: x[0] * (1 if x[2] == 1 else 10000000.0))
bob_pay_time = 0
bob_liked_books = list()
for book in books:
if len(bob_liked_books) >= k:
break
if book[2] == 1 and book[1] == 0:
bob_pay_time += book[0]
bob_liked_books.append(book[3])
books = sorted(
books, key=lambda x: x[0] * (1 if x[1] == 1 and x[2] == 1 else 10000000.0)
)
pay_time = alice_pay_time + bob_pay_time
alice_liked_book_count = len(alice_liked_books)
bob_liked_book_count = len(bob_liked_books)
for book in books:
if book[1] == 1 and book[2] == 1:
if alice_liked_book_count < k or bob_liked_book_count < k:
pay_time += book[0]
alice_liked_book_count += 1
bob_liked_book_count += 1
if alice_liked_book_count > k:
pay_time -= books_map[alice_liked_books[-1]][0]
alice_liked_books.pop()
alice_liked_book_count = k
if bob_liked_book_count > k:
pay_time -= books_map[bob_liked_books[-1]][0]
bob_liked_books.pop()
bob_liked_book_count = k
elif len(alice_liked_books) == 0 or len(bob_liked_books) == 0:
break
elif (
book[0]
< books_map[alice_liked_books[-1]][0] + books_map[bob_liked_books[-1]][0]
):
pay_time -= books_map[alice_liked_books[-1]][0]
pay_time -= books_map[bob_liked_books[-1]][0]
pay_time += book[0]
alice_liked_books.pop()
bob_liked_books.pop()
if alice_liked_book_count >= k and bob_liked_book_count >= k:
print(pay_time)
else:
print(-1) | IMPORT ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR NUMBER NUMBER NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF FUNC_CALL VAR VAR VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR NUMBER NUMBER NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF FUNC_CALL VAR VAR VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER IF VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR IF VAR VAR VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER IF VAR NUMBER BIN_OP VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER VAR VAR VAR NUMBER NUMBER VAR VAR VAR NUMBER NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | def answer():
if n3 + n1 < k:
return -1
if n3 + n2 < k:
return -1
ap = [0]
for i in range(n1):
ap.append(ap[-1] + a[i])
ap.append(0)
bp = [0]
for i in range(n2):
bp.append(bp[-1] + b[i])
bp.append(0)
start = max(max(0, k - n1), max(0, k - n2))
s = 0
for i in range(start):
s += common[i]
common.append(0)
ans = 10000000000.0
for i in range(start, min(k, n3) + 1):
ans = min(ans, s + ap[k - i] + bp[k - i])
s += common[i]
return ans
n, k = map(int, input().split())
a, b, common = [], [], []
for i in range(n):
t, x, y = map(int, input().split())
if x and y:
common.append(t)
elif x == 1 and y == 0:
a.append(t)
elif x == 0 and y == 1:
b.append(t)
common.sort()
a.sort()
b.sort()
n1, n2, n3 = len(a), len(b), len(common)
print(answer()) | FUNC_DEF IF BIN_OP VAR VAR VAR RETURN NUMBER IF BIN_OP VAR VAR VAR RETURN NUMBER ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR RETURN VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
a, b, c = [], [], []
for i in range(n):
t, x, y = map(int, input().split())
if x == 1 and y == 1:
a.append(t)
if x == 1 and y == 0:
b.append(t)
if x == 0 and y == 1:
c.append(t)
a.sort()
b.sort()
c.sort()
prea, preb, prec = 0, 0, 0
cnt = min(len(b), len(c), k)
for i in range(cnt):
preb += b[i]
prec += c[i]
bestans = int(2000000000.0 + 5)
if cnt == k:
bestans = preb + prec
for both in range(min(len(a), k)):
prea += a[both]
idx = k - both - 1
if idx < cnt:
preb -= b[idx]
prec -= c[idx]
if idx - 1 < cnt and prea + preb + prec < bestans:
bestans = prea + preb + prec
print(bestans) if bestans <= int(2000000000.0) else print(-1) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP NUMBER NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR VAR IF BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR EXPR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR FUNC_CALL VAR NUMBER |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
a = []
b = []
ab = []
c = 0
p = 0
e = 0
for q in range(n):
x, y, z = map(int, input().split())
if y == 1 and z == 0:
a.append(x)
elif y == 0 and z == 1:
b.append(x)
elif y == 1 and z == 1:
ab.append(x)
a.sort()
b.sort()
ab.sort()
if len(ab) + min(len(a), len(b)) < k:
print("-1")
else:
for h in range(k):
if p == min(len(a), len(b)):
c = c + sum(ab[e : e + k - h])
break
elif e == len(ab):
c = c + sum(a[p : p + k - h]) + sum(b[p : p + k - h])
break
c = c + min(a[p] + b[p], ab[e])
if a[p] + b[p] < ab[e]:
p = p + 1
else:
e = e + 1
print(c) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR IF VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR VAR IF BIN_OP VAR VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | from sys import stdin
n, k = map(int, input().split())
x = []
y = []
z = []
k1 = 0
k2 = 0
for i in range(n):
t, a, b = map(int, stdin.readline().split())
if a == 1 and b == 1:
x.append(t)
k1 += 1
k2 += 1
elif a == 1:
k1 += 1
y.append(t)
elif b == 1:
k2 += 1
z.append(t)
if k1 < k or k2 < k:
print(-1)
else:
x.sort()
y.sort()
z.sort()
ans = 0
k1 = 0
k2 = 0
p1 = 0
p2 = 0
p3 = 0
lx = len(x)
ly = len(y)
lz = len(z)
for i in range(k):
if p1 >= lx:
ans += y[p2] + z[p3]
p2 += 1
p3 += 1
elif p2 >= ly or p3 >= lz:
ans += x[p1]
p1 += 1
elif x[p1] <= y[p2] + z[p3]:
ans += x[p1]
p1 += 1
else:
ans += y[p2] + z[p3]
p2 += 1
p3 += 1
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR VAR NUMBER IF VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = [int(i) for i in input().split()]
a_likes = list()
b_likes = list()
both_like = list()
a_likes.append(0)
b_likes.append(0)
both_like.append(0)
for book in range(n):
t, a, b = [int(i) for i in input().split()]
if a == 1 and b == 1:
both_like.append(t)
elif a == 1:
a_likes.append(t)
elif b == 1:
b_likes.append(t)
a_likes.sort()
b_likes.sort()
both_like.sort()
for i in range(1, len(a_likes)):
a_likes[i] += a_likes[i - 1]
for i in range(1, len(b_likes)):
b_likes[i] += b_likes[i - 1]
for i in range(1, len(both_like)):
both_like[i] += both_like[i - 1]
if len(both_like) + len(a_likes) < k or len(both_like) + len(b_likes) < k:
print("-1")
else:
ans = 2 * pow(10, 9) + 1
rbo = 0
while rbo < min(len(both_like), k + 1):
if k - rbo < len(a_likes) and k - rbo < len(b_likes):
ans = min(ans, both_like[rbo] + a_likes[k - rbo] + b_likes[k - rbo])
rbo += 1
if ans < 2 * pow(10, 9) + 1:
print(ans)
else:
print("-1") | ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP NUMBER FUNC_CALL VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR NUMBER IF VAR BIN_OP BIN_OP NUMBER FUNC_CALL VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
books = [list(map(int, input().split())) for i in range(n)]
AB, A, B = [[10**5, 0, 0]], [[10**5, 0, 0]], [[10**5, 0, 0]]
for t, a, b in books:
if a == 1 and b == 1:
AB.append([t, a, b])
elif a == 1 and b == 0:
A.append([t, a, b])
elif a == 0 and b == 1:
B.append([t, a, b])
AB = list(reversed(sorted(AB)))
A = list(reversed(sorted(A)))
B = list(reversed(sorted(B)))
ans = []
for i in range(k):
if len(AB) != 1 and AB[-1][0] < A[-1][0] + B[-1][0]:
ans.append(AB.pop()[0])
elif len(A) != 1 and len(B) != 1:
ans.append(A.pop()[0])
ans.append(B.pop()[0])
else:
print(-1)
break
else:
print(sum(ans)) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR LIST LIST BIN_OP NUMBER NUMBER NUMBER NUMBER LIST LIST BIN_OP NUMBER NUMBER NUMBER NUMBER LIST LIST BIN_OP NUMBER NUMBER NUMBER NUMBER FOR VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR LIST VAR VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR LIST VAR VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR LIST VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
a = []
b = []
c = []
at = 0
bt = 0
alo = 0
blo = 0
clo = 0
for i in range(n):
t, al, bl = map(int, input().split())
if al == 1 and bl == 1:
at += 1
bt += 1
clo += 1
c.append(t)
elif al == 1:
at += 1
alo += 1
a.append(t)
elif bl == 1:
bt += 1
blo += 1
b.append(t)
if at < k or bt < k:
print("-1")
else:
total = 0
time = 0
a.sort()
b.sort()
c.sort()
j = 0
r = 0
while total < k:
if j == alo or j == blo:
time = time + c[r]
r += 1
elif r == clo:
time = time + a[j] + b[j]
j += 1
elif a[j] + b[j] < c[r]:
time = time + a[j] + b[j]
j += 1
else:
time = time + c[r]
r += 1
total += 1
print(time) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR NUMBER IF BIN_OP VAR VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
t = []
a = []
b = []
n1 = 0
n2 = 0
n3 = 0
for _ in range(n):
x, y, z = map(int, input().split())
if y == 1 and z == 1:
t.append(x)
n1 += 1
elif y == 1:
a.append(x)
n2 += 1
elif z == 1:
b.append(x)
n3 += 1
t.sort()
a.sort()
b.sort()
nc = min(n2, n3)
c = [(a[i] + b[i]) for i in range(min(n2, n3))]
ans = 0
i = 0
j = 0
while i < n1 and j < nc and k > 0:
if t[i] <= c[j]:
i += 1
ans += t[i - 1]
else:
j += 1
ans += c[j - 1]
k -= 1
if k == 0:
break
while k > 0 and i < n1:
i += 1
ans += t[i - 1]
k -= 1
while k > 0 and j < nc:
j += 1
ans += c[j - 1]
k -= 1
if k > 0:
print(-1)
else:
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR VAR VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER WHILE VAR NUMBER VAR VAR VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR NUMBER WHILE VAR NUMBER VAR VAR VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = [int(x) for x in input().split(" ")]
time = []
a = []
b = []
for i in range(n):
p, q, r = [int(x) for x in input().split(" ")]
time.append(p)
a.append(q)
b.append(r)
oa = []
ob = []
bo = []
no = []
for i in range(n):
if a[i] and b[i]:
bo.append(i)
elif a[i] and b[i] == 0:
oa.append(i)
elif a[i] == 0 and b[i]:
ob.append(i)
else:
no.append(i)
if len(bo) + len(oa) < k or len(bo) + len(ob) < k:
print(-1)
else:
def time_req(i):
return time[i]
oa.sort(key=time_req)
ob.sort(key=time_req)
bo.sort(key=time_req)
oa.reverse()
ob.reverse()
bo.reverse()
books = set()
for i in range(k):
if len(oa) == 0 or len(ob) == 0:
books.add(bo.pop())
continue
elif len(bo) == 0:
books.add(oa.pop())
books.add(ob.pop())
continue
if time[bo[-1]] < time[oa[-1]] + time[ob[-1]] or len(oa) == 0 or len(ob) == 0:
books.add(bo.pop())
else:
books.add(oa.pop())
books.add(ob.pop())
print(sum(list(map(time_req, list(books))))) | ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER FUNC_DEF RETURN VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR IF VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
arr = [[int(i) for i in input().split()] for _ in range(n)]
a = sorted([arr[i][0] for i in range(n) if arr[i][1] == 1 and arr[i][2] == 0])
b = sorted([arr[i][0] for i in range(n) if arr[i][1] == 0 and arr[i][2] == 1])
ab = sorted([arr[i][0] for i in range(n) if arr[i][1] == 1 and arr[i][2] == 1])
ans = 0
l, r = 0, 0
for i in range(k):
v1 = 10**9
if len(a) > l and len(b) > l:
v1 = a[l] + b[l]
v2 = 10**9
if len(ab) > r:
v2 = ab[r]
if v1 == v2 == 10**9:
ans = -1
break
if v1 < v2:
l += 1
ans += v1
else:
r += 1
ans += v2
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP NUMBER NUMBER IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP NUMBER NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR IF VAR VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
group = [[], [], [], []]
for i in range(n):
t, a, b = map(int, input().split())
group[a * 2 + b].append(t)
pref_sum = [[0], [0], [0], [0]]
for g, ps in zip(group, pref_sum):
g.sort()
for i in g:
ps.append(ps[-1] + i)
ans = 10**18
for i in range(min(k, len(group[3])) + 1):
rest = k - i
if rest > len(group[2]) or rest > len(group[1]):
continue
ans = min(ans, pref_sum[3][i] + pref_sum[1][rest] + pref_sum[2][rest])
if ans < 10**18:
print(ans)
else:
print(-1) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST LIST LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR VAR ASSIGN VAR LIST LIST NUMBER LIST NUMBER LIST NUMBER LIST NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FOR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR IF VAR BIN_OP NUMBER NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
both = []
a = []
b = []
for _ in range(n):
t, ai, bi = map(int, input().split())
if ai == 1 and bi == 1:
both.append(t)
elif ai == 1:
a.append(t)
elif bi == 1:
b.append(t)
both.sort()
a.sort()
b.sort()
for l in [both, a, b]:
if len(l) < k:
l += [10**10] * (k - len(l))
if len(a) > k:
a = a[:k]
if len(b) > k:
b = b[:k]
if len(both) > k:
both = both[:k]
out = 0
for i in range(k):
out += min(a[i] + b[i], both[-i - 1])
if out >= 10**10:
print(-1)
else:
print(out) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR LIST VAR VAR VAR IF FUNC_CALL VAR VAR VAR VAR BIN_OP LIST BIN_OP NUMBER NUMBER BIN_OP VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR BIN_OP VAR NUMBER IF VAR BIN_OP NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
a, b, tog = [], [], []
for i in range(n):
t, x, y = map(int, input().split())
if x == 1 and y == 1:
tog.append(t)
elif x == 1 and y == 0:
a.append(t)
elif x == 0 and y == 1:
b.append(t)
ltog = len(tog)
la = len(a)
lb = len(b)
if ltog + la < k or ltog + lb < k:
print(-1)
else:
a.sort(reverse=True)
b.sort(reverse=True)
tog.sort(reverse=True)
ca, cb = 0, 0
res = 0
while ca < k or cb < k:
if tog:
ptog = tog[-1]
else:
ptog = 0
if a:
pa = a[-1]
else:
pa = 0
if b:
pb = b[-1]
else:
pb = 0
if ca < k and cb < k:
if pa and pb and ptog:
if pa + pb < ptog:
res += pa + pb
a.pop()
b.pop()
else:
res += ptog
tog.pop()
ca += 1
cb += 1
elif ptog:
res += ptog
tog.pop()
ca += 1
cb += 1
elif pa and pb:
res += pa + pb
a.pop()
b.pop()
ca += 1
cb += 1
else:
res = -1
break
elif ca < k:
if pa and ptog:
if pa < ptog:
res += pa
a.pop()
ca += 1
else:
res += ptog
tog.pop()
ca += 1
cb += 1
elif ptog:
res += ptog
tog.pop()
ca += 1
cb += 1
elif pa:
res += pa
a.pop()
ca += 1
else:
res = -1
break
elif cb < k:
if pb and ptog:
if pb < ptog:
res += pb
b.pop()
cb += 1
else:
res += ptog
tog.pop()
ca += 1
cb += 1
elif ptog:
res += ptog
tog.pop()
ca += 1
cb += 1
elif pb:
res += pb
b.pop()
cb += 1
else:
res = -1
break
print(res) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR IF VAR ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR VAR IF VAR VAR VAR IF BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR IF VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR IF VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | k, n = [int(i) for i in input().split()]
a = []
b = []
both = []
for _ in range(k):
t, x, y = [int(i) for i in input().split()]
if x == 1 and y == 1:
both.append(t)
elif x == 1:
a.append(t)
elif y == 1:
b.append(t)
a.sort()
b.sort()
both.sort()
if len(a) + len(both) < n or len(b) + len(both) < n:
print(-1)
quit()
bI = 0
aI = 0
bothI = 0
count = 0
t = 0
while count < n:
count += 1
if len(a) != aI and len(b) != bI and len(both) != bothI:
if a[aI] + b[bI] < both[bothI]:
t += a[aI] + b[bI]
aI += 1
bI += 1
else:
t += both[bothI]
bothI += 1
elif len(a) != aI and len(b) != bI:
t += a[aI] + b[bI]
aI += 1
bI += 1
else:
t += both[bothI]
bothI += 1
print(t) | ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | class Book:
def __init__(self, time, for_alice, for_bob):
self.time = time
self.for_alice = bool(for_alice)
self.for_bob = bool(for_bob)
self.status = ("A" if self.for_alice else "") + ("B" if self.for_bob else "")
def get(li, i):
try:
return li[i].time
except IndexError:
return 10**10
n, k = map(int, input().split())
books = [Book(*map(int, input().split())) for _ in range(n)]
alice_rest, bob_rest = k, k
for_alice_books = list(filter(lambda x: x.status == "A", books))
for_bob_books = list(filter(lambda x: x.status == "B", books))
for_both_books = list(filter(lambda x: x.status == "AB", books))
for_alice_books = sorted(for_alice_books, key=lambda x: x.time)
for_bob_books = sorted(for_bob_books, key=lambda x: x.time)
for_both_books = sorted(for_both_books, key=lambda x: x.time)
for_alice_books_index, for_bob_books_index, for_both_books_index = 0, 0, 0
total = 0
while alice_rest > 0 or bob_rest > 0:
alice_var = get(for_alice_books, for_alice_books_index)
bob_var = get(for_bob_books, for_bob_books_index)
both_var = get(for_both_books, for_both_books_index)
if all(i == 10**10 for i in [alice_var, bob_var, both_var]):
total = -1
break
if alice_rest > 0 and bob_rest > 0:
if alice_var + bob_var > both_var:
for_both_books_index += 1
total += both_var
else:
for_alice_books_index += 1
for_bob_books_index += 1
total += alice_var + bob_var
alice_rest -= 1
bob_rest -= 1
elif alice_rest == 0:
if bob_var > both_var:
for_both_books_index += 1
total += both_var
else:
for_bob_books_index += 1
total += bob_var
bob_rest -= 1
else:
if alice_var > both_var:
for_both_books_index += 1
total += both_var
else:
for_alice_books_index += 1
total += alice_var
alice_rest -= 1
print(total if total < 10**10 else -1) | CLASS_DEF FUNC_DEF ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR STRING STRING VAR STRING STRING FUNC_DEF RETURN VAR VAR VAR RETURN BIN_OP NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR STRING VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR STRING VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR STRING VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR BIN_OP NUMBER NUMBER VAR LIST VAR VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER IF BIN_OP VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR VAR NUMBER VAR NUMBER IF VAR NUMBER IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP NUMBER NUMBER VAR NUMBER |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
both, f, s = [], [], []
for i in range(n):
t, x, y = map(int, input().split())
if x == 1 and y == 1:
both.append(t)
elif x == 1:
f.append(t)
elif y == 1:
s.append(t)
if min(len(f), len(s)) + len(both) < k:
print(-1)
else:
both = sorted(both)
f = sorted(f)
s = sorted(s)
g = [both, f, s]
for l in g:
for j, v in enumerate(l):
if j > 0:
l[j] = l[j - 1] + l[j]
if len(f) > len(s):
b = f
f = s
s = b
mn = 5 * 10**9
if len(f) == 0:
mn = both[k - 1]
for i in range(len(f) + 1):
if i > k:
break
elif i > 0:
if i == k:
mn = min(f[i - 1] + s[i - 1], mn)
elif len(both) >= k - i:
mn = min(f[i - 1] + s[i - 1] + both[k - i - 1], mn)
elif i == 0:
if len(both) >= k:
mn = both[k - 1]
print(mn) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR VAR VAR FOR VAR VAR FOR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP NUMBER BIN_OP NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR IF VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR IF FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER VAR IF VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | t, k = map(int, input().split())
A = []
B = []
C = []
for i in range(t):
a, b, c = map(int, input().split())
if b == 1 and c == 0:
A.append(a)
elif b == 0 and c == 1:
B.append(a)
elif b == 1:
C.append(a)
if min(len(A), len(B)) + len(C) < k:
print(-1)
else:
A.sort()
B.sort()
s = 0
for i in range(min(len(A), len(B))):
C.append(A[i] + B[i])
C.sort()
for i in range(k):
s += C[i]
print(s) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | import sys
def solve():
n, k = map(int, sys.stdin.readline().split())
list_book = [[], [], []]
count = [0, 0, 0]
for _ in range(n):
t, a, b = map(int, sys.stdin.readline().split())
if a and b:
list_book[0].append(t)
count[0] += 1
elif a:
list_book[1].append(t)
count[1] += 1
elif b:
list_book[2].append(t)
count[2] += 1
if count[0] + min(count[1], count[2]) < k:
print(-1)
return
list_book[0].sort(reverse=True)
list_book[1].sort(reverse=True)
list_book[2].sort(reverse=True)
result = 0
book = k
result = 0
for _ in range(k):
next_time = 1000000000000000000000
token = 0
if list_book[1] and list_book[2]:
next_time = list_book[1][-1] + list_book[2][-1]
if list_book[0]:
if next_time > list_book[0][-1]:
next_time = list_book[0][-1]
token = 1
if token:
list_book[0].pop()
else:
list_book[1].pop()
list_book[2].pop()
result += next_time
print(result)
solve() | IMPORT FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST LIST LIST LIST ASSIGN VAR LIST NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER VAR VAR NUMBER NUMBER IF VAR EXPR FUNC_CALL VAR NUMBER VAR VAR NUMBER NUMBER IF VAR EXPR FUNC_CALL VAR NUMBER VAR VAR NUMBER NUMBER IF BIN_OP VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER RETURN EXPR FUNC_CALL VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER IF VAR NUMBER IF VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, K = list(map(int, input().split()))
l = [list(map(int, input().split())) for _ in range(n)]
al, bo, bt = [], [], []
for i in l:
if i[1] == 1 and i[2] == 1:
bt.append(i[0])
elif i[1] == 1 and i[2] == 0:
al.append(i[0])
elif i[1] == 0 and i[2] == 1:
bo.append(i[0])
al.sort()
bt.sort()
bo.sort()
i, j, k = 0, 0, 0
ans, cnt = 0, 0
while i < len(al) and j < len(bo) and k < len(bt) and cnt < K:
if al[i] + bo[j] < bt[k]:
ans += al[i] + bo[j]
i += 1
j += 1
else:
ans += bt[k]
k += 1
cnt += 1
while k < len(bt) and cnt < K:
ans += bt[k]
k += 1
cnt += 1
while cnt < K and i < len(al) and j < len(bo):
ans += al[i] + bo[j]
i += 1
j += 1
cnt += 1
if cnt < K:
print(-1)
else:
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR IF BIN_OP VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER WHILE VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
li = [[list(), list()], [list(), list()]]
for i in range(n):
a, b, c = map(int, input().split())
li[b][c].append(a)
cnt = 0
itr1, itr2, itr3 = 0, 0, 0
li[0][1].sort()
li[1][0].sort()
li[1][1].sort()
ttl = 0
run = True
for i in range(k):
sm = 0
if itr1 != len(li[0][1]) and itr2 != len(li[1][0]):
sm = li[0][1][itr1] + li[1][0][itr2]
if itr3 != len(li[1][1]):
if sm < li[1][1][itr3]:
itr1 += 1
itr2 += 1
ttl += sm
else:
ttl += li[1][1][itr3]
itr3 += 1
else:
itr1 += 1
itr2 += 1
ttl += sm
elif itr3 != len(li[1][1]):
ttl += li[1][1][itr3]
itr3 += 1
else:
run = False
break
if run:
print(ttl)
else:
print(-1) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST LIST FUNC_CALL VAR FUNC_CALL VAR LIST FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER NUMBER VAR IF VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR NUMBER NUMBER VAR VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER NUMBER VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR VAR IF VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
alice = []
bob = []
both = []
counta = 0
countb = 0
for i in range(0, n):
l1 = list(map(int, input().split()))
if l1[1] == 1 and l1[2] == 1:
counta += 1
countb += 1
both.append(l1)
elif l1[1] == 1:
counta += 1
alice.append(l1)
elif l1[2] == 1:
countb += 1
bob.append(l1)
if counta < k or countb < k:
print(-1)
else:
t = 0
alice.sort()
bob.sort()
both.sort()
a = 0
b = 0
for i in range(1, k + 1):
if b > min(len(alice), len(bob)) - 1:
for j in range(a, a + k - i + 1):
t = t + both[j][0]
break
if a > len(both) - 1:
for j in range(b, b + k - i + 1):
t = t + alice[j][0] + bob[j][0]
break
if both[a][0] <= alice[b][0] + bob[b][0]:
t = t + both[a][0]
a = a + 1
else:
t = t + alice[b][0] + bob[b][0]
b = b + 1
print(t) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR NUMBER VAR VAR NUMBER IF VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | from sys import maxsize
n, k = map(int, input().split())
l = [[0], [0], [0], [0]]
x = []
ans = maxsize
for i in range(n):
x.append(list(map(int, input().split())))
x.sort()
for p in x:
temp = l[p[1] * 2 + p[2]][-1]
l[p[1] * 2 + p[2]].append(temp + p[0])
for i in range(min(k + 1, len(l[3]))):
if k - i < len(l[1]) and k - i < len(l[2]):
ans = min(ans, l[3][i] + l[1][k - i] + l[2][k - i])
if ans == maxsize:
print(-1)
else:
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST LIST NUMBER LIST NUMBER LIST NUMBER LIST NUMBER ASSIGN VAR LIST ASSIGN VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR VAR ASSIGN VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER BIN_OP VAR VAR VAR NUMBER BIN_OP VAR VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | [N, K] = list(map(int, input().split()))
L = [[[] for i in range(2)] for j in range(2)]
for i in range(N):
[t, a, b] = list(map(int, input().split()))
L[a][b].append(t)
INF = int(1000000000000.0)
for i in range(2):
for j in range(2):
L[i][j].sort()
for k in range(K - len(L[i][j])):
L[i][j].append(INF)
ans = INF
now = 0
for i in range(K):
now += L[1][1][i]
ans = min(ans, now)
for i in range(K):
now -= L[1][1][K - i - 1]
now += L[1][0][i]
now += L[0][1][i]
ans = min(ans, now)
if ans >= INF:
ans = -1
print(ans) | ASSIGN LIST VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN LIST VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER BIN_OP BIN_OP VAR VAR NUMBER VAR VAR NUMBER NUMBER VAR VAR VAR NUMBER NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | import sys
def input():
return sys.stdin.readline().strip()
def list2d(a, b, c):
return [([c] * b) for i in range(a)]
def list3d(a, b, c, d):
return [[([d] * c) for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e):
return [[[([e] * d) for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1):
return int(-(-x // y))
def INT():
return int(input())
def MAP():
return map(int, input().split())
def LIST(N=None):
return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes():
print("Yes")
def No():
print("No")
def YES():
print("YES")
def NO():
print("NO")
INF = 10**19
MOD = 10**9 + 7
N, K = MAP()
A = [INF]
B = [INF]
AB = [INF]
for i in range(N):
t, a, b = MAP()
if a == b == 1:
AB.append(t)
elif a == 1:
A.append(t)
elif b == 1:
B.append(t)
A.sort(reverse=1)
B.sort(reverse=1)
AB.sort(reverse=1)
ans = 0
for i in range(K):
if AB[-1] <= A[-1] + B[-1]:
ans += AB.pop()
else:
ans += A.pop() + B.pop()
if ans >= INF:
print(-1)
exit()
print(ans) | IMPORT FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN BIN_OP LIST VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF RETURN BIN_OP LIST VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF RETURN BIN_OP LIST VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF NUMBER RETURN FUNC_CALL VAR BIN_OP VAR VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF NONE RETURN VAR NONE FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF EXPR FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR LIST VAR ASSIGN VAR LIST VAR ASSIGN VAR LIST VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
t10 = []
t01 = []
t11 = []
for x in range(n):
t, a, b = map(int, input().split())
if a == 0 and b == 1:
t01.append(t)
elif a == 1 and b == 0:
t10.append(t)
elif a == 1 and b == 1:
t11.append(t)
t10.sort()
t01.sort()
t11.sort()
for x in range(1, len(t01)):
t01[x] += t01[x - 1]
for x in range(1, len(t10)):
t10[x] += t10[x - 1]
for x in range(1, len(t11)):
t11[x] += t11[x - 1]
t10 = [0] + t10
t01 = [0] + t01
t11 = [0] + t11
ans = []
for x in range(len(t11)):
if k - x > len(t10) - 1 or k - x > len(t01) - 1:
continue
t = 0
t += t11[x]
t += t01[k - x]
t += t10[k - x]
ans.append(t)
if x == k:
break
if len(ans) == 0:
print(-1)
else:
print(min(ans)) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = map(int, input().split())
common, alice, bob = [0], [0], [0]
for i in range(n):
t, a, b = map(int, input().split())
if a + b == 2:
common.append(t)
elif a == 1:
alice.append(t)
elif b == 1:
bob.append(t)
common.sort()
alice.sort()
bob.sort()
for i in range(1, len(common)):
common[i] += common[i - 1]
for i in range(1, len(alice)):
alice[i] += alice[i - 1]
for i in range(1, len(bob)):
bob[i] += bob[i - 1]
ans = 10**20
for i in range(len(common)):
t = common[i]
r = k - i
if r >= 0 and r < len(alice) and r < len(bob):
t += alice[r] + bob[r]
ans = min(ans, t)
if ans < 10**20:
print(ans)
else:
print(-1) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR LIST NUMBER LIST NUMBER LIST NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR BIN_OP NUMBER NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | from sys import stdin
input = stdin.readline
n, k = map(int, input().split())
al = []
bl = []
abl = []
for _ in range(n):
t, a, b = map(int, input().split())
if a and b:
abl.append(t)
elif a:
al.append(t)
elif b:
bl.append(t)
al.sort()
bl.sort()
abl.sort()
sl = list(map(lambda o: o[0] + o[1], zip(al, bl)))
abl.reverse()
sl.reverse()
c = 0
while abl and sl and k:
if abl[-1] < sl[-1]:
c += abl.pop()
else:
c += sl.pop()
k -= 1
if k:
if len(abl) + len(sl) < k:
print(-1)
else:
while k:
if abl:
c += abl.pop()
else:
c += sl.pop()
k -= 1
print(c)
else:
print(c) | ASSIGN VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR VAR VAR IF VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER IF VAR IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER WHILE VAR IF VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | line = input().split()
n, k = int(line[0]), int(line[1])
a, b, c = [], [], []
for i in range(n):
line = input().split()
t, ai, bi = int(line[0]), int(line[1]), int(line[2])
if ai == 1 and bi == 1:
c.append(t)
elif ai == 1:
a.append(t)
elif bi == 1:
b.append(t)
a.sort()
b.sort()
c.sort()
i, j, cnt = 0, 0, 0
ans = 0
while i < min(len(a), len(b)) and j < len(c) and cnt < k:
if a[i] + b[i] < c[j]:
ans += a[i] + b[i]
i += 1
else:
ans += c[j]
j += 1
cnt += 1
while i < min(len(a), len(b)) and cnt < k:
ans += a[i] + b[i]
cnt += 1
i += 1
while j < len(c) and cnt < k:
ans += c[j]
cnt += 1
j += 1
if cnt < k:
print(-1)
else:
print(ans) | ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR IF BIN_OP VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER WHILE VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | seg = [0] * 200000
def offset(x):
return x + 100000
def upd(node, L, R, pos, val):
if L + 1 == R:
seg[node] += val
seg[offset(node)] = seg[node] * L
return
M = (L + R) // 2
if pos < M:
upd(node << 1, L, M, pos, val)
else:
upd(node << 1 | 1, M, R, pos, val)
seg[node] = seg[node << 1] + seg[node << 1 | 1]
seg[offset(node)] = seg[offset(node << 1)] + seg[offset(node << 1 | 1)]
def query(node, L, R, k):
if k == 0:
return [0, 0]
if seg[node] < k:
return [seg[offset(node)], seg[node]]
if L + 1 == R:
return [L * k, k]
M = (L + R) // 2
left = query(node << 1, L, M, k)
right = query(node << 1 | 1, M, R, k - left[1])
left[0] += right[0]
left[1] += right[1]
return left
n, m, k = map(int, input().split())
A, B, both, neither = [], [], [], []
for i in range(n):
t, a, b = map(int, input().split())
if a == 0 and b == 0:
neither.append([t, i + 1])
if a == 1 and b == 0:
A.append([t, i + 1])
if a == 0 and b == 1:
B.append([t, i + 1])
if a == 1 and b == 1:
both.append([t, i + 1])
upd(1, 0, 10001, t, 1)
A.sort()
B.sort()
both.sort()
p1 = min(k, len(both))
p2 = k - p1
if 2 * k - p1 > m or p2 > min(len(A), len(B)):
print(-1)
exit(0)
sum, ans, ch = 0, 2**31, p1
for i in range(p1):
sum += both[i][0]
upd(1, 0, 10001, both[i][0], -1)
for i in range(p2):
sum += A[i][0] + B[i][0]
upd(1, 0, 10001, A[i][0], -1)
upd(1, 0, 10001, B[i][0], -1)
ans = sum + query(1, 0, 10001, m - 2 * k + p1)[0]
while p1 > 0:
if p2 == min(len(A), len(B)):
break
upd(1, 0, 10001, A[p2][0], -1)
sum += A[p2][0]
upd(1, 0, 10001, B[p2][0], -1)
sum += B[p2][0]
upd(1, 0, 10001, both[p1 - 1][0], 1)
sum -= both[p1 - 1][0]
p2 += 1
p1 -= 1
if m - 2 * k + p1 < 0:
break
Q = query(1, 0, 10001, m - 2 * k + p1)
if ans > sum + Q[0]:
ans = sum + Q[0]
ch = p1
print(ans)
ind = (
[both[i][1] for i in range(ch)]
+ [A[i][1] for i in range(k - ch)]
+ [B[i][1] for i in range(k - ch)]
)
st = (
neither
+ [both[i] for i in range(ch, len(both))]
+ [A[i] for i in range(k - ch, len(A))]
+ [B[i] for i in range(k - ch, len(B))]
)
st.sort()
ind += [st[i][1] for i in range(m - 2 * k + ch)]
print(" ".join([str(x) for x in ind])) | ASSIGN VAR BIN_OP LIST NUMBER NUMBER FUNC_DEF RETURN BIN_OP VAR NUMBER FUNC_DEF IF BIN_OP VAR NUMBER VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR RETURN ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_DEF IF VAR NUMBER RETURN LIST NUMBER NUMBER IF VAR VAR VAR RETURN LIST VAR FUNC_CALL VAR VAR VAR VAR IF BIN_OP VAR NUMBER VAR RETURN LIST BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR BIN_OP VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER RETURN VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR LIST LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR LIST VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR LIST VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR LIST VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR LIST VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR IF BIN_OP BIN_OP NUMBER VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR NUMBER BIN_OP NUMBER NUMBER VAR FOR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR FUNC_CALL VAR NUMBER NUMBER NUMBER BIN_OP BIN_OP VAR BIN_OP NUMBER VAR VAR NUMBER WHILE VAR NUMBER IF VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER VAR VAR NUMBER NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER VAR VAR NUMBER NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR NUMBER IF BIN_OP BIN_OP VAR BIN_OP NUMBER VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER NUMBER BIN_OP BIN_OP VAR BIN_OP NUMBER VAR VAR IF VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP NUMBER VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = list(map(int, input().split()))
time, A, B, AB, lenA, lenB, lenAB = [], [], [], [], 0, 0, 0
for idx in range(n):
t, a, b = list(map(int, input().split()))
time.append(t)
if a and not b:
A.append(t)
lenA += 1
elif b and not a:
B.append(t)
lenB += 1
elif a and b:
AB.append(t)
lenAB += 1
if lenA + lenAB < k or lenB + lenAB < k:
print(-1)
else:
A.sort()
B.sort()
AB.sort()
s, x, y = 0, 0, 0
for i in range(k):
if x < lenA and x < lenB and y < lenAB:
if A[x] + B[x] < AB[y]:
s += A[x] + B[x]
x += 1
else:
s += AB[y]
y += 1
elif x >= lenA or x >= lenB:
s += AB[y]
y += 1
elif y >= lenAB and x < lenA and x < lenB:
s += A[x] + B[x]
x += 1
print(s) | ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR LIST LIST LIST LIST NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER IF BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR IF BIN_OP VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but... Their mom doesn't think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are $n$ books in the family library. The $i$-th book is described by three integers: $t_i$ — the amount of time Alice and Bob need to spend to read it, $a_i$ (equals $1$ if Alice likes the $i$-th book and $0$ if not), and $b_i$ (equals $1$ if Bob likes the $i$-th book and $0$ if not).
So they need to choose some books from the given $n$ books in such a way that:
Alice likes at least $k$ books from the chosen set and Bob likes at least $k$ books from the chosen set; the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it's shared between them) and they read all books together, so the total reading time is the sum of $t_i$ over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
-----Input-----
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$).
The next $n$ lines contain descriptions of books, one description per line: the $i$-th line contains three integers $t_i$, $a_i$ and $b_i$ ($1 \le t_i \le 10^4$, $0 \le a_i, b_i \le 1$), where:
$t_i$ — the amount of time required for reading the $i$-th book; $a_i$ equals $1$ if Alice likes the $i$-th book and $0$ otherwise; $b_i$ equals $1$ if Bob likes the $i$-th book and $0$ otherwise.
-----Output-----
If there is no solution, print only one integer -1. Otherwise print one integer $T$ — the minimum total reading time of the suitable set of books.
-----Examples-----
Input
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
Output
18
Input
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
Output
8
Input
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
Output
-1 | n, k = [int(i) for i in input().split()]
doub = []
a = []
b = []
for i in range(n):
ti, ai, bi = [int(i) for i in input().split()]
if ai and bi:
doub.append(ti)
elif ai:
a.append(ti)
elif bi:
b.append(ti)
doub.sort()
a.sort()
b.sort()
time = 0
i = 0
j = 0
while i + j < k:
if i < len(doub):
db = doub[i]
else:
db = 100000
if j < len(a) and j < len(b):
sing = a[j] + b[j]
else:
sing = 100000
if db <= sing:
time += db
i += 1
else:
time += sing
j += 1
if len(doub) + min(len(a), len(b)) >= k:
print(time)
else:
print(-1) | ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR VAR IF VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER |
Bessie is out grazing on the farm, which consists of $n$ fields connected by $m$ bidirectional roads. She is currently at field $1$, and will return to her home at field $n$ at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has $k$ special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field $1$ to field $n$. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
-----Input-----
The first line contains integers $n$, $m$, and $k$ ($2 \le n \le 2 \cdot 10^5$, $n-1 \le m \le 2 \cdot 10^5$, $2 \le k \le n$) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains $k$ integers $a_1, a_2, \ldots, a_k$ ($1 \le a_i \le n$) — the special fields. All $a_i$ are distinct.
The $i$-th of the following $m$ lines contains integers $x_i$ and $y_i$ ($1 \le x_i, y_i \le n$, $x_i \ne y_i$), representing a bidirectional road between fields $x_i$ and $y_i$.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
-----Output-----
Output one integer, the maximum possible length of the shortest path from field $1$ to $n$ after Farmer John installs one road optimally.
-----Examples-----
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
-----Note-----
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields $3$ and $5$, and the resulting shortest path from $1$ to $5$ is length $3$.
The graph for the second example is shown below. Farmer John must add a road between fields $2$ and $4$, and the resulting shortest path from $1$ to $5$ is length $3$. | def main():
n, m, k = [int(s) for s in input().strip().split()]
A = [int(s) for s in input().strip().split()]
adj = [[] for _ in range(n + 1)]
for _ in range(m):
u, v = [int(s) for s in input().strip().split()]
adj[u].append(v)
adj[v].append(u)
def bfs(u):
result = [float("inf")] * (n + 1)
result[u] = 0
step = 0
q = [u]
while q:
step += 1
newq = []
for u in q:
for v in adj[u]:
if result[v] == float("inf"):
result[v] = step
newq.append(v)
q = newq
return result
X = bfs(1)
Y = bfs(n)
A.sort(key=lambda a: X[a] - Y[a])
result, maxx = float("-inf"), float("-inf")
for a in A:
result = max(result, maxx + Y[a])
maxx = max(maxx, X[a])
result = min(X[n], result + 1)
print(result)
main() | FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR FUNC_DEF ASSIGN VAR BIN_OP LIST FUNC_CALL VAR STRING BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST VAR WHILE VAR VAR NUMBER ASSIGN VAR LIST FOR VAR VAR FOR VAR VAR VAR IF VAR VAR FUNC_CALL VAR STRING ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR STRING FUNC_CALL VAR STRING FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Bessie is out grazing on the farm, which consists of $n$ fields connected by $m$ bidirectional roads. She is currently at field $1$, and will return to her home at field $n$ at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has $k$ special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field $1$ to field $n$. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
-----Input-----
The first line contains integers $n$, $m$, and $k$ ($2 \le n \le 2 \cdot 10^5$, $n-1 \le m \le 2 \cdot 10^5$, $2 \le k \le n$) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains $k$ integers $a_1, a_2, \ldots, a_k$ ($1 \le a_i \le n$) — the special fields. All $a_i$ are distinct.
The $i$-th of the following $m$ lines contains integers $x_i$ and $y_i$ ($1 \le x_i, y_i \le n$, $x_i \ne y_i$), representing a bidirectional road between fields $x_i$ and $y_i$.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
-----Output-----
Output one integer, the maximum possible length of the shortest path from field $1$ to $n$ after Farmer John installs one road optimally.
-----Examples-----
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
-----Note-----
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields $3$ and $5$, and the resulting shortest path from $1$ to $5$ is length $3$.
The graph for the second example is shown below. Farmer John must add a road between fields $2$ and $4$, and the resulting shortest path from $1$ to $5$ is length $3$. | import sys
input = sys.stdin.readline
n, m, K = map(int, input().split())
sp = list(map(lambda x: int(x) - 1, input().split()))
peer = [[] for _ in range(n)]
for _ in range(m):
a, b = map(int, input().split())
a -= 1
b -= 1
peer[a].append(b)
peer[b].append(a)
a0 = [(10**6) for _ in range(n)]
a0[0] = 0
a1 = [(10**6) for _ in range(n)]
a1[-1] = 0
now = [0]
while now:
last = now
now = []
for x in last:
for y in peer[x]:
if a0[y] > a0[x] + 1:
a0[y] = a0[x] + 1
now.append(y)
now = [n - 1]
while now:
last = now
now = []
for x in last:
for y in peer[x]:
if a1[y] > a1[x] + 1:
a1[y] = a1[x] + 1
now.append(y)
xyxy = []
for w in sp:
xyxy.append([a0[w] - a1[w], a0[w], a1[w]])
xyxy.sort()
xx = []
maxi = 0
for j in range(K):
xx.append(max(maxi, xyxy[j][1]))
can = []
for j in range(1, K):
can.append(xx[j - 1] + xyxy[j][2] + 1)
print(min(a0[-1], max(can))) | IMPORT ASSIGN VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP NUMBER NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER NUMBER ASSIGN VAR LIST NUMBER WHILE VAR ASSIGN VAR VAR ASSIGN VAR LIST FOR VAR VAR FOR VAR VAR VAR IF VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST BIN_OP VAR NUMBER WHILE VAR ASSIGN VAR VAR ASSIGN VAR LIST FOR VAR VAR FOR VAR VAR VAR IF VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR LIST BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR |
Bessie is out grazing on the farm, which consists of $n$ fields connected by $m$ bidirectional roads. She is currently at field $1$, and will return to her home at field $n$ at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has $k$ special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field $1$ to field $n$. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
-----Input-----
The first line contains integers $n$, $m$, and $k$ ($2 \le n \le 2 \cdot 10^5$, $n-1 \le m \le 2 \cdot 10^5$, $2 \le k \le n$) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains $k$ integers $a_1, a_2, \ldots, a_k$ ($1 \le a_i \le n$) — the special fields. All $a_i$ are distinct.
The $i$-th of the following $m$ lines contains integers $x_i$ and $y_i$ ($1 \le x_i, y_i \le n$, $x_i \ne y_i$), representing a bidirectional road between fields $x_i$ and $y_i$.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
-----Output-----
Output one integer, the maximum possible length of the shortest path from field $1$ to $n$ after Farmer John installs one road optimally.
-----Examples-----
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
-----Note-----
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields $3$ and $5$, and the resulting shortest path from $1$ to $5$ is length $3$.
The graph for the second example is shown below. Farmer John must add a road between fields $2$ and $4$, and the resulting shortest path from $1$ to $5$ is length $3$. | import sys
input = sys.stdin.readline
N, M, K = map(int, input().split())
A = list(map(lambda x: int(x) - 1, input().split()))
graph = [[] for _ in range(N)]
for _ in range(M):
a, b = map(int, input().split())
graph[a - 1].append(b - 1)
graph[b - 1].append(a - 1)
def bfs(s):
D = [-1] * N
D[s] = 0
q = [s]
while q:
qq = []
for p in q:
for np in graph[p]:
if D[np] == -1:
D[np] = D[p] + 1
qq.append(np)
q = qq
return D
D1 = bfs(0)
D2 = bfs(N - 1)
distance = D1[N - 1]
P = []
for a in A:
d1 = D1[a]
d2 = D2[a]
P.append((d1, d2))
P.sort()
ans = 0
for i in range(K - 1):
b1, b2 = P[i]
c1, c2 = P[i + 1]
ans = max(ans, min(c1 + b2 + 1, c2 + b1 + 1))
ans = min(ans, distance)
print(ans) | IMPORT ASSIGN VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR VAR NUMBER ASSIGN VAR LIST VAR WHILE VAR ASSIGN VAR LIST FOR VAR VAR FOR VAR VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR LIST FOR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Bessie is out grazing on the farm, which consists of $n$ fields connected by $m$ bidirectional roads. She is currently at field $1$, and will return to her home at field $n$ at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has $k$ special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field $1$ to field $n$. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
-----Input-----
The first line contains integers $n$, $m$, and $k$ ($2 \le n \le 2 \cdot 10^5$, $n-1 \le m \le 2 \cdot 10^5$, $2 \le k \le n$) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains $k$ integers $a_1, a_2, \ldots, a_k$ ($1 \le a_i \le n$) — the special fields. All $a_i$ are distinct.
The $i$-th of the following $m$ lines contains integers $x_i$ and $y_i$ ($1 \le x_i, y_i \le n$, $x_i \ne y_i$), representing a bidirectional road between fields $x_i$ and $y_i$.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
-----Output-----
Output one integer, the maximum possible length of the shortest path from field $1$ to $n$ after Farmer John installs one road optimally.
-----Examples-----
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
-----Note-----
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields $3$ and $5$, and the resulting shortest path from $1$ to $5$ is length $3$.
The graph for the second example is shown below. Farmer John must add a road between fields $2$ and $4$, and the resulting shortest path from $1$ to $5$ is length $3$. | def bfs(adj, root):
q = [(root, 0)]
dist = [0] * len(adj)
vis = {root}
i = 0
while i < len(q):
u, d = q[i]
i += 1
dist[u] = d
for v in adj[u]:
if v not in vis:
vis.add(v)
q.append((v, d + 1))
return dist
def sol(n, sfs, edges):
adj = [[] for _ in range(n + 1)]
for u, v in edges:
adj[u].append(v)
adj[v].append(u)
dista = bfs(adj, 1)
distb = bfs(adj, n)
orig = distb[1]
ans = 0
sfs.sort(key=lambda sf: dista[sf] - distb[sf])
maxdista = dista[sfs[0]]
for fs in sfs[1:]:
ans = max(ans, maxdista + distb[fs])
maxdista = max(maxdista, dista[fs])
return min(orig, ans + 1)
n, m, _ = map(int, input().split())
sfs = list(map(int, input().split()))
edges = [tuple(map(int, input().split())) for _ in range(m)]
print(sol(n, sfs, edges)) | FUNC_DEF ASSIGN VAR LIST VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR FOR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER FOR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR |
Baby Ehab has a piece of Cut and Stick with an array a of length n written on it. He plans to grab a pair of scissors and do the following to it:
* pick a range (l, r) and cut out every element a_l, a_{l + 1}, ..., a_r in this range;
* stick some of the elements together in the same order they were in the array;
* end up with multiple pieces, where every piece contains some of the elements and every element belongs to some piece.
More formally, he partitions the sequence a_l, a_{l + 1}, ..., a_r into subsequences. He thinks a partitioning is beautiful if for every piece (subsequence) it holds that, if it has length x, then no value occurs strictly more than ⌈ x/2 ⌉ times in it.
He didn't pick a range yet, so he's wondering: for q ranges (l, r), what is the minimum number of pieces he needs to partition the elements a_l, a_{l + 1}, ..., a_r into so that the partitioning is beautiful.
A sequence b is a subsequence of an array a if b can be obtained from a by deleting some (possibly zero) elements. Note that it does not have to be contiguous.
Input
The first line contains two integers n and q (1 ≤ n,q ≤ 3 ⋅ 10^5) — the length of the array a and the number of queries.
The second line contains n integers a_1, a_2, ..., a_{n} (1 ≤ a_i ≤ n) — the elements of the array a.
Each of the next q lines contains two integers l and r (1 ≤ l ≤ r ≤ n) — the range of this query.
Output
For each query, print the minimum number of subsequences you need to partition this range into so that the partitioning is beautiful. We can prove such partitioning always exists.
Example
Input
6 2
1 3 2 3 3 2
1 6
2 5
Output
1
2
Note
In the first query, you can just put the whole array in one subsequence, since its length is 6, and no value occurs more than 3 times in it.
In the second query, the elements of the query range are [3,2,3,3]. You can't put them all in one subsequence, since its length is 4, and 3 occurs more than 2 times. However, you can partition it into two subsequences: [3] and [2,3,3]. | import sys
input = sys.stdin.readline
n, q = map(int, input().split())
counts = [0] * (n + 1)
c_amt = [0] * (n + 3)
c_amt[0] = n + 1
max_count = 0
moves = 0
def add(x):
global max_count, moves
moves += 1
v = l[x]
c_amt[counts[v]] -= 1
counts[v] += 1
c_amt[counts[v]] += 1
max_count = max(counts[v], max_count)
def remove(x):
global max_count, moves
moves += 1
v = l[x]
c_amt[counts[v]] -= 1
if max_count == counts[v] and c_amt[counts[v]] == 0:
max_count -= 1
counts[v] -= 1
c_amt[counts[v]] += 1
l = list(map(int, input().split()))
left = 0
right = -1
out = [0] * q
queries = []
for i in range(q):
ll, rr = map(int, input().split())
queries.append((ll - 1, rr - 1, i))
k = 547
queries.sort(key=lambda x: 2 * n * (x[0] // k) + x[1] * (-1) ** (x[0] // k))
for ll, rr, i in queries:
while right < rr:
add(right + 1)
right += 1
while right > rr:
remove(right)
right -= 1
while left > ll:
add(left - 1)
left -= 1
while left < ll:
remove(left)
left += 1
sz = rr - ll + 1
top = max_count
other = sz - top
out[i] = max(1, top - other)
print("\n".join(map(str, out))) | IMPORT ASSIGN VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FUNC_DEF VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR NUMBER VAR VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_DEF VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP NUMBER BIN_OP VAR NUMBER VAR FOR VAR VAR VAR VAR WHILE VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER WHILE VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER WHILE VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER WHILE VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR |
Given a char array representing tasks CPU need to do. It contains capital letters A to Z where different letters represent different tasks.Tasks could be done without original order. Each task could be done in one interval. For each interval, CPU could finish one task or just be idle.
However, there is a non-negative cooling interval n that means between two same tasks, there must be at least n intervals that CPU are doing different tasks or just be idle.
You need to return the least number of intervals the CPU will take to finish all the given tasks.
Example 1:
Input: tasks = ["A","A","A","B","B","B"], n = 2
Output: 8
Explanation: A -> B -> idle -> A -> B -> idle -> A -> B.
Note:
The number of tasks is in the range [1, 10000].
The integer n is in the range [0, 100]. | class Solution:
def leastInterval(self, tasks, n):
l, dic = len(tasks), {}
for c in tasks:
if c in dic:
dic[c] += 1
else:
dic[c] = 1
m, a = max(dic.values()), 0
for c in dic:
if dic[c] == m:
a += 1
return max(l, (m - 1) * (n + 1) + a) | CLASS_DEF FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR DICT FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER FOR VAR VAR IF VAR VAR VAR VAR NUMBER RETURN FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR |
Given a char array representing tasks CPU need to do. It contains capital letters A to Z where different letters represent different tasks.Tasks could be done without original order. Each task could be done in one interval. For each interval, CPU could finish one task or just be idle.
However, there is a non-negative cooling interval n that means between two same tasks, there must be at least n intervals that CPU are doing different tasks or just be idle.
You need to return the least number of intervals the CPU will take to finish all the given tasks.
Example 1:
Input: tasks = ["A","A","A","B","B","B"], n = 2
Output: 8
Explanation: A -> B -> idle -> A -> B -> idle -> A -> B.
Note:
The number of tasks is in the range [1, 10000].
The integer n is in the range [0, 100]. | class Solution:
def leastInterval(self, tasks, n):
if not tasks:
return 0
counts = {}
for i in tasks:
if i in counts:
counts[i] += 1
else:
counts[i] = 1
M = max(counts.values())
Mct = sum([(1) for i in counts if counts[i] == M])
return max(len(tasks), (M - 1) * (n + 1) + Mct) | CLASS_DEF FUNC_DEF IF VAR RETURN NUMBER ASSIGN VAR DICT FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR NUMBER VAR VAR VAR VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR |
Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment l_{i} and ends at moment r_{i}.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all n shows. Are two TVs enough to do so?
-----Input-----
The first line contains one integer n (1 ≤ n ≤ 2·10^5) — the number of shows.
Each of the next n lines contains two integers l_{i} and r_{i} (0 ≤ l_{i} < r_{i} ≤ 10^9) — starting and ending time of i-th show.
-----Output-----
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
-----Examples-----
Input
3
1 2
2 3
4 5
Output
YES
Input
4
1 2
2 3
2 3
1 2
Output
NO | h = int(input())
programas = []
tvs = [-1, -1]
for i in range(h):
programas.append(list(map(int, input().split())))
programas.sort()
teste = True
for i in programas:
horario = min(tvs)
if i[0] <= horario:
print("NO")
teste = False
break
else:
tvs[tvs.index(horario)] = i[1]
if teste:
print("YES") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING |
Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment l_{i} and ends at moment r_{i}.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all n shows. Are two TVs enough to do so?
-----Input-----
The first line contains one integer n (1 ≤ n ≤ 2·10^5) — the number of shows.
Each of the next n lines contains two integers l_{i} and r_{i} (0 ≤ l_{i} < r_{i} ≤ 10^9) — starting and ending time of i-th show.
-----Output-----
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
-----Examples-----
Input
3
1 2
2 3
4 5
Output
YES
Input
4
1 2
2 3
2 3
1 2
Output
NO | n = int(input())
a = []
for i in range(n):
a.append(list(map(int, input().split())))
a.sort()
b = a[0][1]
c = -1
e = 0
for i in range(1, n):
if a[i][0] > b:
b = a[i][1]
elif a[i][0] > c:
c = a[i][1]
else:
e = 1
break
if e == 1:
print("NO")
else:
print("YES") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF VAR VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment l_{i} and ends at moment r_{i}.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all n shows. Are two TVs enough to do so?
-----Input-----
The first line contains one integer n (1 ≤ n ≤ 2·10^5) — the number of shows.
Each of the next n lines contains two integers l_{i} and r_{i} (0 ≤ l_{i} < r_{i} ≤ 10^9) — starting and ending time of i-th show.
-----Output-----
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
-----Examples-----
Input
3
1 2
2 3
4 5
Output
YES
Input
4
1 2
2 3
2 3
1 2
Output
NO | from sys import stdin
input = stdin.readline
q = []
counter = 0
n = int(input())
for i in range(n):
s, e = map(int, input().split())
q.append([s, True])
q.append([e + 1, False])
q.sort()
for i in range(len(q)):
if q[i][1]:
counter += 1
else:
counter -= 1
if counter >= 3:
print("NO")
exit()
print("YES") | ASSIGN VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR LIST VAR NUMBER EXPR FUNC_CALL VAR LIST BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING |
Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment l_{i} and ends at moment r_{i}.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all n shows. Are two TVs enough to do so?
-----Input-----
The first line contains one integer n (1 ≤ n ≤ 2·10^5) — the number of shows.
Each of the next n lines contains two integers l_{i} and r_{i} (0 ≤ l_{i} < r_{i} ≤ 10^9) — starting and ending time of i-th show.
-----Output-----
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
-----Examples-----
Input
3
1 2
2 3
4 5
Output
YES
Input
4
1 2
2 3
2 3
1 2
Output
NO | n = int(input())
vec = []
for _ in range(n):
a, b = [int(x) for x in input().split()]
vec.append([a, b])
vec.sort()
f1 = f2 = 0
flag1 = flag2 = True
flag = "YES"
for item in vec:
if f1 <= f2:
if flag1:
f1 = item[1]
flag1 = False
elif item[0] > f1:
f1 = item[1]
else:
flag = "NO"
break
elif flag2:
f2 = item[1]
flag2 = False
elif item[0] > f2:
f2 = item[1]
else:
flag = "NO"
break
print(flag) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR LIST VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR STRING FOR VAR VAR IF VAR VAR IF VAR ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR ASSIGN VAR VAR NUMBER ASSIGN VAR STRING IF VAR ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR ASSIGN VAR VAR NUMBER ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR |
Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment l_{i} and ends at moment r_{i}.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all n shows. Are two TVs enough to do so?
-----Input-----
The first line contains one integer n (1 ≤ n ≤ 2·10^5) — the number of shows.
Each of the next n lines contains two integers l_{i} and r_{i} (0 ≤ l_{i} < r_{i} ≤ 10^9) — starting and ending time of i-th show.
-----Output-----
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
-----Examples-----
Input
3
1 2
2 3
4 5
Output
YES
Input
4
1 2
2 3
2 3
1 2
Output
NO | n = int(input())
a = [list(map(int, input().split())) for i in range(n)]
a.sort()
p = -1
q = -1
f = 0
for l, r in a:
if l > p:
p = r
elif l > q:
q = r
else:
f = 1
break
if f:
print("NO")
else:
print("YES") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment l_{i} and ends at moment r_{i}.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all n shows. Are two TVs enough to do so?
-----Input-----
The first line contains one integer n (1 ≤ n ≤ 2·10^5) — the number of shows.
Each of the next n lines contains two integers l_{i} and r_{i} (0 ≤ l_{i} < r_{i} ≤ 10^9) — starting and ending time of i-th show.
-----Output-----
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
-----Examples-----
Input
3
1 2
2 3
4 5
Output
YES
Input
4
1 2
2 3
2 3
1 2
Output
NO | shows = int(input())
tv1 = []
tv2 = []
horarios = []
for i in range(shows):
horario = [*map(int, input().split())]
horarios.append(horario)
horarios.sort(key=lambda x: x[0])
possible = "YES"
for horario in horarios:
if len(tv1) == 0:
tv1.append(horario)
elif len(tv2) == 0:
tv2.append(horario)
elif tv1[-1][1] < horario[0]:
tv1.append(horario)
elif tv2[-1][1] < horario[0]:
tv2.append(horario)
else:
possible = "NO"
break
print(possible) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR STRING FOR VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR |
Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment l_{i} and ends at moment r_{i}.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all n shows. Are two TVs enough to do so?
-----Input-----
The first line contains one integer n (1 ≤ n ≤ 2·10^5) — the number of shows.
Each of the next n lines contains two integers l_{i} and r_{i} (0 ≤ l_{i} < r_{i} ≤ 10^9) — starting and ending time of i-th show.
-----Output-----
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
-----Examples-----
Input
3
1 2
2 3
4 5
Output
YES
Input
4
1 2
2 3
2 3
1 2
Output
NO | n = int(input())
shows = []
MAX_TVS = 2
for i in range(n):
show_info = [int(x) for x in input().split()]
shows.append(show_info)
shows.sort(key=lambda x: x[0])
tvs = []
i = 0
while len(tvs) <= MAX_TVS and i < n:
if len(tvs) == MAX_TVS:
if shows[i][0] > tvs[0][1]:
tvs.pop(0)
elif shows[i][0] > tvs[1][1]:
tvs.pop(1)
tvs.append(shows[i])
i += 1
result = "YES"
if len(tvs) > MAX_TVS:
result = "NO"
print(result) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE FUNC_CALL VAR VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR STRING IF FUNC_CALL VAR VAR VAR ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR |
Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment l_{i} and ends at moment r_{i}.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all n shows. Are two TVs enough to do so?
-----Input-----
The first line contains one integer n (1 ≤ n ≤ 2·10^5) — the number of shows.
Each of the next n lines contains two integers l_{i} and r_{i} (0 ≤ l_{i} < r_{i} ≤ 10^9) — starting and ending time of i-th show.
-----Output-----
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
-----Examples-----
Input
3
1 2
2 3
4 5
Output
YES
Input
4
1 2
2 3
2 3
1 2
Output
NO | n = int(input())
x = []
for i in range(n):
l, r = map(int, input().split())
x.append((l, 1))
x.append((r, -1))
x.sort(key=lambda x: (x[0], -x[1]))
s = 0
for i in range(len(x)):
s += x[i][1]
if s == 3:
print("NO")
exit()
print("YES") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING |
Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment l_{i} and ends at moment r_{i}.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all n shows. Are two TVs enough to do so?
-----Input-----
The first line contains one integer n (1 ≤ n ≤ 2·10^5) — the number of shows.
Each of the next n lines contains two integers l_{i} and r_{i} (0 ≤ l_{i} < r_{i} ≤ 10^9) — starting and ending time of i-th show.
-----Output-----
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
-----Examples-----
Input
3
1 2
2 3
4 5
Output
YES
Input
4
1 2
2 3
2 3
1 2
Output
NO | n = int(input())
v = []
for i in range(n):
s = input()
s = [int(x) for x in s.split()]
v.append(s)
v.sort()
w = []
last = -1
for i in range(n):
if v[i][0] <= last:
w.append(v[i])
else:
last = v[i][1]
last = -1
f = 1
for i in range(len(w)):
if w[i][0] <= last:
f = 0
break
else:
last = w[i][1]
if f:
print("YES")
else:
print("NO") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment l_{i} and ends at moment r_{i}.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all n shows. Are two TVs enough to do so?
-----Input-----
The first line contains one integer n (1 ≤ n ≤ 2·10^5) — the number of shows.
Each of the next n lines contains two integers l_{i} and r_{i} (0 ≤ l_{i} < r_{i} ≤ 10^9) — starting and ending time of i-th show.
-----Output-----
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
-----Examples-----
Input
3
1 2
2 3
4 5
Output
YES
Input
4
1 2
2 3
2 3
1 2
Output
NO | n = int(input())
shows = []
for i in range(n):
l, r = map(int, input().split())
shows.append((l, r))
shows.sort()
a_endtime, b_endtime = -1, -1
for show in shows:
if show[0] <= a_endtime:
print("NO")
break
else:
a_endtime = show[1]
if a_endtime > b_endtime:
a_endtime, b_endtime = b_endtime, a_endtime
else:
print("YES") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR IF VAR NUMBER VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING |
Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment l_{i} and ends at moment r_{i}.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all n shows. Are two TVs enough to do so?
-----Input-----
The first line contains one integer n (1 ≤ n ≤ 2·10^5) — the number of shows.
Each of the next n lines contains two integers l_{i} and r_{i} (0 ≤ l_{i} < r_{i} ≤ 10^9) — starting and ending time of i-th show.
-----Output-----
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
-----Examples-----
Input
3
1 2
2 3
4 5
Output
YES
Input
4
1 2
2 3
2 3
1 2
Output
NO | def main():
n = int(input())
lst = list()
for i in range(n):
line = input()
l, r = map(int, line.split())
lst.append((l, r))
lst.sort()
tv1 = [(-1, -1)]
tv2 = [(-1, -1)]
for i in range(n):
if lst[i][0] > tv1[-1][1]:
tv1.append(lst[i])
elif lst[i][0] > tv2[-1][1]:
tv2.append(lst[i])
tv1.remove(tv1[0])
tv2.remove(tv2[0])
tv = tv1 + tv2
tv.sort()
if lst != tv:
return "NO"
else:
return "YES"
rs = main()
print(rs) | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR IF VAR VAR RETURN STRING RETURN STRING ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR |
Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment l_{i} and ends at moment r_{i}.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all n shows. Are two TVs enough to do so?
-----Input-----
The first line contains one integer n (1 ≤ n ≤ 2·10^5) — the number of shows.
Each of the next n lines contains two integers l_{i} and r_{i} (0 ≤ l_{i} < r_{i} ≤ 10^9) — starting and ending time of i-th show.
-----Output-----
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
-----Examples-----
Input
3
1 2
2 3
4 5
Output
YES
Input
4
1 2
2 3
2 3
1 2
Output
NO | n = int(input())
l = []
for _ in range(n):
l.append(tuple(map(int, input().split())))
l.sort()
t1 = -1
t2 = -1
for i in l:
if i[0] > t1:
t1 = i[1]
elif i[0] > t2:
t2 = i[1]
else:
print("NO")
exit()
print("YES") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR ASSIGN VAR VAR NUMBER IF VAR NUMBER VAR ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING |
Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment l_{i} and ends at moment r_{i}.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all n shows. Are two TVs enough to do so?
-----Input-----
The first line contains one integer n (1 ≤ n ≤ 2·10^5) — the number of shows.
Each of the next n lines contains two integers l_{i} and r_{i} (0 ≤ l_{i} < r_{i} ≤ 10^9) — starting and ending time of i-th show.
-----Output-----
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
-----Examples-----
Input
3
1 2
2 3
4 5
Output
YES
Input
4
1 2
2 3
2 3
1 2
Output
NO | l = []
t = int(input())
for _ in range(t):
n, m = map(int, input().split())
l.append((n, m))
l.sort()
le, ri = -1, -1
for i in range(t):
if l[i][0] > le:
le = l[i][1]
elif l[i][0] > ri:
ri = l[i][1]
else:
print("NO")
exit()
print("YES") | ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF VAR VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING |
Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment l_{i} and ends at moment r_{i}.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all n shows. Are two TVs enough to do so?
-----Input-----
The first line contains one integer n (1 ≤ n ≤ 2·10^5) — the number of shows.
Each of the next n lines contains two integers l_{i} and r_{i} (0 ≤ l_{i} < r_{i} ≤ 10^9) — starting and ending time of i-th show.
-----Output-----
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
-----Examples-----
Input
3
1 2
2 3
4 5
Output
YES
Input
4
1 2
2 3
2 3
1 2
Output
NO | import sys
n = int(input())
a = []
tv1 = -1
tv2 = -1
for _ in range(n):
a += [tuple(map(int, sys.stdin.readline().split()))]
a.sort()
for i in range(n):
s, e = a[i]
if s > tv1:
tv1 = e
elif s > tv2:
tv2 = e
else:
break
else:
print("YES")
exit()
print("NO") | IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR LIST FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING |
Polycarp is a great fan of television.
He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment l_{i} and ends at moment r_{i}.
Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.
Polycarp wants to check out all n shows. Are two TVs enough to do so?
-----Input-----
The first line contains one integer n (1 ≤ n ≤ 2·10^5) — the number of shows.
Each of the next n lines contains two integers l_{i} and r_{i} (0 ≤ l_{i} < r_{i} ≤ 10^9) — starting and ending time of i-th show.
-----Output-----
If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).
-----Examples-----
Input
3
1 2
2 3
4 5
Output
YES
Input
4
1 2
2 3
2 3
1 2
Output
NO | def main():
n = int(input())
events = []
for _ in range(n):
l, r = (int(x) for x in input().split())
events.append((l, 1))
events.append((r + 1, -1))
cur = 0
for _, event in sorted(events):
cur += event
if cur > 2:
print("NO")
break
else:
print("YES")
main() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.