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 | n, k = list(map(int, input().split()))
ca, cb = 0, 0
x, y, z = [], [], []
for i in range(n):
t, a, b = list(map(int, input().split()))
ca += a
cb += b
if a == 1 and b == 0:
x.append(t)
elif a == 0 and b == 1:
y.append(t)
elif a == 1 and b == 1:
z.append(t)
x.sort()
y.sort()
z.sort()
if ca < k or cb < k:
print("-1")
else:
total = 0
arr = []
for i in range(min(len(z), k)):
total += z[i]
arr.append(z[i])
if len(z) < k:
for i in range(len(z), k):
total += x[0] + y[0]
del x[0]
del y[0]
while len(x) > 0 and len(y) > 0 and len(arr) > 0 and x[0] + y[0] < arr[-1]:
total += x[0] + y[0] - arr[-1]
del arr[-1]
del x[0]
del y[0]
print(total) | ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR 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 IF VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER WHILE FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER 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 | n, k = map(int, input().split())
a = []
b = []
ab = []
for i in range(n):
t, a1, b1 = map(int, input().split())
if a1 == 1 and b1 == 1:
ab.append(t)
elif a1 == 1 and b1 == 0:
a.append(t)
elif a1 == 0 and b1 == 1:
b.append(t)
a.sort()
a_i = 0
b.sort()
ab.sort()
ab_i = 0
sum_t = 0
count = 0
while count < k:
if (a_i == len(a) or a_i == len(b)) and ab_i < len(ab):
sum_t += ab[ab_i]
ab_i += 1
elif ab_i == len(ab) and a_i < len(a) and a_i < len(b):
sum_t += a[a_i] + b[a_i]
a_i += 1
elif ab_i < len(ab) and a_i < len(a) and a_i < len(b):
if a[a_i] + b[a_i] < ab[ab_i]:
sum_t += a[a_i] + b[a_i]
a_i += 1
else:
sum_t += ab[ab_i]
ab_i += 1
else:
sum_t = -1
break
count += 1
print(sum_t) | 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 ASSIGN VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER IF 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 VAR VAR VAR NUMBER ASSIGN 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())
ab = []
a = []
b = []
for i in range(n):
t, c, d = map(int, input().split())
if c and d == 0:
a.append(t)
elif d and c == 0:
b.append(t)
elif c * d:
ab.append(t)
a.sort()
b.sort()
ab.sort()
la = len(a)
lb = len(b)
lab = len(ab)
if la + lab < k or lb + lab < k:
print(-1)
exit()
if lb > la:
la, lb = lb, la
a, b = b, a
ans = 0
if lab >= k:
ans += sum(ab[:k])
now = k - 1
na = 0
nb = 0
while nb < lb and now >= 0 and ab[now] > a[na] + b[nb]:
ans -= ab[now] - a[na] - b[nb]
na += 1
nb += 1
now -= 1
else:
ans += sum(ab) + sum(b[: k - lab]) + sum(a[: k - lab])
now = lab - 1
na = k - lab
nb = k - lab
while nb < lb and now >= 0 and ab[now] > a[na] + b[nb]:
ans -= ab[now] - a[na] - b[nb]
na += 1
nb += 1
now -= 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 NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP VAR VAR EXPR FUNC_CALL VAR 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 VAR VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR NUMBER VAR VAR BIN_OP VAR VAR VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR WHILE VAR VAR VAR NUMBER VAR VAR BIN_OP VAR VAR VAR VAR VAR BIN_OP BIN_OP VAR VAR 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, k = map(int, input().split())
same = []
different_a = []
different_b = []
for i in range(n):
t, x, y = map(int, input().split())
if x == 1 and y == 1:
same.append(t)
elif x == 1 and y == 0:
different_a.append(t)
elif y == 1 and x == 0:
different_b.append(t)
flag = 1
if len(same) + len(different_a) < k or len(same) + len(different_b) < k:
flag = 0
if flag == 0:
print(-1)
else:
same.sort()
for i in range(1, len(same)):
same[i] += same[i - 1]
different_a.sort()
for i in range(1, len(different_a)):
different_a[i] += different_a[i - 1]
different_b.sort()
for i in range(1, len(different_b)):
different_b[i] += different_b[i - 1]
i = 0
minn = 10**15
while i <= k:
ans1 = 10**15
if i - 1 >= 0 and i - 1 < len(same):
ans1 = same[i - 1]
ans2 = 10**15
if k - i - 1 >= 0 and k - i - 1 < len(different_a):
ans2 = different_a[k - i - 1]
ans3 = 10**15
if k - i - 1 >= 0 and k - i - 1 < len(different_b):
ans3 = different_b[k - i - 1]
ans = 0
if i == 0:
ans += ans2 + ans3
elif i == k:
ans += ans1
else:
ans += ans1 + ans2 + ans3
minn = min(ans, minn)
i += 1
print(minn) | 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 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 ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP NUMBER NUMBER IF BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER NUMBER BIN_OP BIN_OP VAR VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER NUMBER BIN_OP BIN_OP VAR VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR IF VAR VAR VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL 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())
inf = float("inf")
a, b, c = [inf], [inf], [inf]
a_cnt, b_cnt = 0, 0
for i in range(n):
u, v, w = map(int, input().split())
if v == 1 and w == 1:
c.append(u)
a_cnt += 1
b_cnt += 1
elif v == 1:
a.append(u)
a_cnt += 1
elif w == 1:
b.append(u)
b_cnt += 1
if a_cnt < k or b_cnt < k:
print(-1)
exit()
a.sort(reverse=True)
b.sort(reverse=True)
c.sort(reverse=True)
a_left, b_left, ans = k, k, 0
while a_left > 0 or b_left > 0:
if a_left and b_left and len(a) - 1 and len(b) - 1 and len(c) - 1:
if a[-1] + b[-1] >= c[-1] or len(a) - 1 < a_left or len(b) - 1 < b_left:
ans += c.pop()
else:
ans += a.pop() + b.pop()
a_left -= 1
b_left -= 1
else:
best = inf
if a_left and b_left:
best = min(best, a[-1] + b[-1])
if a_left and not b_left:
best = min(best, a[-1])
if b_left and not a_left:
best = min(best, b[-1])
best = min(best, c[-1])
if best == c[-1]:
ans += c.pop()
a_left -= 1
b_left -= 1
elif best == a[-1]:
ans += a.pop()
a_left -= 1
else:
ans += b.pop()
b_left -= 1
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR VAR VAR LIST VAR LIST VAR LIST VAR 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 EXPR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR 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 ASSIGN VAR VAR VAR VAR VAR NUMBER WHILE VAR NUMBER VAR NUMBER IF VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER IF VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL 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 = list(map(int, input().split(" ")))
db = []
a = []
b = []
for i in range(n):
t, A, B = list(map(int, input().split(" ")))
if A == B == 1:
db.append(t)
elif A == 1:
a.append(t)
elif B == 1:
b.append(t)
a.sort()
b.sort()
for i in range(min(len(a), len(b))):
db.append(a[i] + b[i])
if len(db) < k:
print(-1)
else:
db.sort()
print(sum(db[:k])) | ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL 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 FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING 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 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 IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER 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
I = lambda: list(map(int, input().split()))
n, k = I()
a = []
b = []
ab = []
for _ in range(n):
x, y, z = I()
if y and z:
ab.append(x)
elif y:
a.append(x)
elif z:
b.append(x)
a.sort()
b.sort()
ab.sort()
x, y, z = len(a), len(b), len(ab)
i = j = l = t = an = 0
if x + z < k or y + z < k:
print(-1)
else:
while t != k:
if i >= x or j >= y:
an += ab[l]
l += 1
elif l >= z:
an += a[i] + b[j]
i += 1
j += 1
elif a[i] + b[j] < ab[l]:
an += a[i] + b[j]
i += 1
j += 1
else:
an += ab[l]
l += 1
t += 1
print(an) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL 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 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 VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER IF BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER WHILE VAR VAR IF VAR VAR VAR VAR VAR VAR VAR 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())
y = []
x = []
z = []
for i in range(n):
a, b, c = map(int, input().split())
if b == 1 and c == 0:
x.append(a)
if c == 1 and b == 0:
y.append(a)
if b == 1 and c == 1:
z.append(a)
x = sorted(x)
y = sorted(y)
z = sorted(z)
for i in range(1, len(x)):
x[i] += x[i - 1]
for i in range(1, len(y)):
y[i] += y[i - 1]
for i in range(1, len(z)):
z[i] += z[i - 1]
if len(x) + len(z) < k or len(y) + len(z) < k:
print(-1)
else:
ans = float("inf")
if len(x) >= k and len(y) >= k:
ans = x[k - 1] + y[k - 1]
if len(z) >= k:
ans = min(z[k - 1], ans)
if x and y:
for i in range(len(z)):
if k - i - 1 - 1 < len(x) and k - i - 1 - 1 < len(y) and k - i + 2 >= 0:
ans = min(ans, z[i] + x[k - i - 2] + y[k - i - 2])
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 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 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 NUMBER ASSIGN VAR FUNC_CALL VAR STRING IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR IF VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP BIN_OP BIN_OP VAR VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER 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 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 = input().split()
n, k = int(n), int(k)
both = []
ali = []
bob = []
while n > 0:
t, a, b = input().split()
t, a, b = int(t), int(a), int(b)
if a == 1 and b == 1:
both.append(t)
elif a == 1:
ali.append(t)
elif b == 1:
bob.append(t)
n -= 1
ans = 0
both.sort(reverse=True)
ali.sort(reverse=True)
bob.sort(reverse=True)
if len(both) + min(len(ali), len(bob)) < k:
print(-1)
else:
while k > 0:
if len(both) == 0 or len(ali) * len(bob) > 0 and ali[-1] + bob[-1] <= both[-1]:
x = ali.pop(-1)
y = bob.pop(-1)
ans += x + y
else:
x = both.pop(-1)
ans += x
k -= 1
print(ans) | ASSIGN VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST WHILE VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR 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 VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL 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 NUMBER WHILE VAR NUMBER IF FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL 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 | import sys
input = lambda: sys.stdin.buffer.readline().rstrip()
q, w = map(int, input().split())
e = []
r = []
t = []
z = 0
x = 0
for i in range(q):
a, b, c = map(int, input().split())
if b == 1 and c == 1:
e.append(a)
z += 1
x += 1
elif b == 1 and c == 0:
r.append(a)
x += 1
elif b == 0 and c == 1:
t.append(a)
z += q
r = sorted(r)
t = sorted(t)
for i in range(min(len(r), len(t))):
e.append(r[i] + t[i])
e = sorted(e)
i = 0
t = 0
k = 0
b = len(e)
if b < w:
print(-1)
else:
while i < w:
if t < b:
k += e[t]
t += 1
i += 1
print(k) | IMPORT ASSIGN VAR FUNC_CALL FUNC_CALL VAR 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 VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR 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 ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER WHILE VAR VAR IF 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 | from sys import stdin, stdout
def input():
return stdin.readline().strip()
N, K = input().split(" ")
N = int(N)
K = int(K)
alice_only = []
bob_only = []
both = []
for _ in range(N):
t, a, b = input().split(" ")
t = int(t)
a = int(a)
b = int(b)
if a == 0 and b == 0:
continue
if a == 1 and b == 1:
both += [t]
if a == 1 and b == 0:
alice_only += [t]
if a == 0 and b == 1:
bob_only += [t]
hybrid = []
alice_only = sorted(alice_only)
bob_only = sorted(bob_only)
for a, b in zip(alice_only, bob_only):
hybrid += [a + b]
candidates = sorted(both + hybrid)
if len(candidates) < K:
print(-1)
else:
print(sum(candidates[0:K])) | FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER VAR LIST VAR IF VAR NUMBER VAR NUMBER VAR LIST VAR IF VAR NUMBER VAR NUMBER VAR LIST VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR FUNC_CALL VAR VAR VAR VAR LIST BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR 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())
a = 0
b = 0
alike = []
blike = []
bothlike = []
for _ in range(n):
t, aa, bb = map(int, input().split())
if aa == 1:
a += 1
if bb == 1:
b += 1
if aa == 1 and bb == 0:
alike.append(t)
elif aa == 0 and bb == 1:
blike.append(t)
elif aa == 1 and bb == 1:
bothlike.append(t)
if a < k or b < k:
print(-1)
else:
alike.sort()
blike.sort()
la = len(alike)
lb = len(blike)
ans = 0
if la < k and lb >= k:
for i in range(la):
bothlike.append(alike[i] + blike[i])
bothlike.sort()
ans = sum(bothlike[:k])
elif lb < k and la >= k:
for i in range(lb):
bothlike.append(alike[i] + blike[i])
bothlike.sort()
ans = sum(bothlike[:k])
elif la >= k and lb >= k:
for i in range(k):
bothlike.append(alike[i] + blike[i])
bothlike.sort()
ans = sum(bothlike[:k])
else:
for i in range(min(la, lb)):
bothlike.append(alike[i] + blike[i])
bothlike.sort()
ans = sum(bothlike[:k])
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER 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 IF 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 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 ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL 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())
c = []
d = []
e = []
for i in range(n):
t, a, b = map(int, input().split())
if a == 1 and b == 0:
c.append(t)
elif b == 1 and a == 0:
d.append(t)
elif a == 1 and b == 1:
e.append(t)
if len(c) + len(e) < k or len(d) + len(e) < k:
print(-1)
else:
c = sorted(c)
d = sorted(d)
e = sorted(e)
l1 = len(c)
l2 = len(d)
l3 = len(e)
s = 0
i1 = 0
i2 = 0
i3 = 0
while i1 + i3 < k or i2 + i3 < k:
if i3 < l3:
if i2 + i3 >= k and i1 < l1:
if e[i3] <= c[i1]:
s += e[i3]
i3 += 1
else:
s += c[i1]
i1 += 1
elif i1 + i3 >= k and i2 < l2:
if e[i3] <= d[i2]:
s += e[i3]
i3 += 1
else:
s += d[i2]
i2 += 1
elif i1 < l1 and i2 < l2:
if e[i3] <= c[i1] + d[i2]:
s += e[i3]
i3 += 1
else:
s += c[i1] + d[i2]
i1 += 1
i2 += 1
else:
s += e[i3]
i3 += 1
else:
s += c[i1] + d[i2]
i1 += 1
i2 += 1
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 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 FUNC_CALL VAR 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 ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR IF VAR VAR IF BIN_OP VAR VAR VAR VAR VAR IF VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER IF BIN_OP VAR VAR VAR VAR VAR IF VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR VAR 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 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, k1 = map(int, input().split())
A = []
B = []
AB = []
for i in range(n):
t, a, b = map(int, input().split())
if a == 1 and b == 1:
AB.append(t)
elif a == 1:
A.append(t)
elif b == 1:
B.append(t)
AB.sort()
A.sort()
B.sort()
if len(AB) + len(A) >= k1 and len(AB) + len(B) >= k1:
a = 0
b = 0
i = 0
j = 0
k = 0
ans = 0
books = 0
while i < len(A) and j < len(B) and k < len(AB):
if A[i] + B[j] <= AB[k]:
ans += A[i] + B[j]
i += 1
j += 1
books += 1
else:
ans += AB[k]
k += 1
books += 1
if books == k1:
break
if i >= len(A) and books != k1:
ans += sum(AB[k : k + k1 - books])
elif j >= len(B) and books != k1:
ans += sum(AB[k : k + k1 - books])
elif k >= len(AB) and books != k1:
ans += sum(A[i : i + k1 - books]) + sum(B[j : j + k1 - books])
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 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 ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE 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 VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR IF VAR FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR IF VAR FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR IF VAR FUNC_CALL 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 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())
at = []
bt = []
both = []
f = 0
for x in range(n):
t, a, b = map(int, input().split())
if a == 1 and b == 1:
both.append(t)
elif a == 1:
at.append(t)
elif b == 1:
bt.append(t)
at.sort()
bt.sort()
both.sort()
i = 0
j = 0
if len(at) + len(both) < k or len(bt) + len(both) < k:
f = -1
else:
for x in range(k):
if i >= len(at) or i >= len(bt):
i = -1
if j >= len(both):
j = -1
if i == -1:
f += both[j]
j += 1
elif j == -1:
f += at[i] + bt[i]
i += 1
elif at[i] + bt[i] < both[j]:
f += at[i] + bt[i]
i += 1
else:
f += both[j]
j += 1
print(f) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR 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 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 NUMBER IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR 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 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 = []
an = k
bn = k
for i in range(n):
t1, a1, b1 = map(int, input().split())
t.append(t1)
a.append(a1)
b.append(b1)
both_like = []
a_like = []
b_like = []
for i in range(n):
if a[i] == 1 and b[i] == 1:
both_like.append(t[i])
elif a[i] == 1 and b[i] == 0:
a_like.append(t[i])
elif a[i] == 0 and b[i] == 1:
b_like.append(t[i])
both_like.sort()
a_like.sort()
b_like.sort()
ans = 0
i = 0
j = 0
k = 0
while i < len(both_like) and an > 0 and bn > 0:
if j < len(a_like) and k < len(b_like) and both_like[i] >= a_like[j] + b_like[k]:
ans += a_like[j] + b_like[k]
j += 1
k += 1
an -= 1
bn -= 1
else:
ans += both_like[i]
i += 1
an -= 1
bn -= 1
while an > 0 and j < len(a_like):
ans += a_like[j]
j += 1
an -= 1
while bn > 0 and k < len(b_like):
ans += b_like[k]
k += 1
bn -= 1
if an == 0 and bn == 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 ASSIGN VAR VAR ASSIGN VAR 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 EXPR FUNC_CALL VAR VAR EXPR 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 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 EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP 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 WHILE VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER WHILE VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR NUMBER 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())
aa, bb, cc = [0], [0], [0]
for _ in range(n):
t, a, b = map(int, input().split())
if a == b and a == 1:
cc.append(t)
elif a == 1:
aa.append(t)
elif b == 1:
bb.append(t)
aa.sort()
bb.sort()
cc.sort()
for i in range(1, len(aa)):
aa[i] = aa[i - 1] + aa[i]
for i in range(1, len(bb)):
bb[i] = bb[i - 1] + bb[i]
for i in range(1, len(cc)):
cc[i] = cc[i - 1] + cc[i]
if len(cc) + len(bb) < k + 2 or len(cc) + len(aa) < k + 2:
print(-1)
else:
ans = float("inf")
for i in range(max(0, k - min(len(aa) - 1, len(bb) - 1)), min(len(cc), k + 1)):
ans = min(ans, cc[i] + bb[k - i] + aa[k - i])
print(ans) | 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 VAR 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 ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER 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 | import sys
inp = [int(x) for x in sys.stdin.read().split()]
ii = 0
n, k = inp[ii : ii + 2]
ii += 2
both, a, b = [], [], []
for i in range(n):
x, p, q = inp[ii : ii + 3]
ii += 3
if p and q:
both.append(x)
elif p:
a.append(x)
elif q:
b.append(x)
if len(both) + len(a) < k or len(both) + len(b) < k:
print(-1)
else:
res = 2 * 10**9
both.sort()
a.sort()
b.sort()
cnt, tot, i, j = len(both), sum(both[0 : min(len(both), k)]), 0, 0
for cnt in range(min(len(both), k), max(k - min(len(a), len(b)), 0) - 1, -1):
while i < k - cnt:
tot += a[i]
i += 1
while j < k - cnt:
tot += b[j]
j += 1
res = min(res, tot)
if cnt > 0:
tot -= both[cnt - 1]
print(res) | IMPORT ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER 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 ASSIGN VAR BIN_OP NUMBER BIN_OP NUMBER NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER NUMBER NUMBER WHILE VAR BIN_OP VAR VAR VAR VAR VAR VAR NUMBER WHILE VAR BIN_OP VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER 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 | line = input()
n, m, k = [int(i) for i in line.split(" ")]
books, allL, aliceL, bobL, other = list(range(1, n + 1)), [], [], [], []
ts = [[] for _ in range(n + 1)]
for i in range(n):
line = input()
t, a, b = [int(j) for j in line.split(" ")]
ts[i + 1] = [t, a, b]
if a == 1 and b == 1:
allL.append(i + 1)
elif a == 1:
aliceL.append(i + 1)
elif b == 1:
bobL.append(i + 1)
else:
other.append(i + 1)
if (
len(allL) + min(len(aliceL), len(bobL)) < k
or len(allL) < k
and 2 * (k - len(allL)) > m - len(allL)
):
print(-1)
exit()
books.sort(key=lambda x: (ts[x][0], -ts[x][2] * ts[x][1]))
allL.sort(key=lambda x: (ts[x][0], -ts[x][2] * ts[x][1]))
aliceL.sort(key=lambda x: (ts[x][0], -ts[x][2] * ts[x][1]))
bobL.sort(key=lambda x: (ts[x][0], -ts[x][2] * ts[x][1]))
other.sort(key=lambda x: (ts[x][0], -ts[x][2] * ts[x][1]))
x = max(
2 * k - m,
0,
k - min(len(aliceL), len(bobL)),
m - len(aliceL) - len(bobL) - len(other),
)
cura, curb, curo, cur = max(0, k - x), max(0, k - x), 0, sum(ts[i][0] for i in allL[:x])
cur += sum(ts[i][0] for i in aliceL[:cura]) + sum(ts[i][0] for i in bobL[:curb])
while cura + x + curb + curo < m:
an = ts[aliceL[cura]][0] if cura < len(aliceL) else 9999999999
bn = ts[bobL[curb]][0] if curb < len(bobL) else 9999999999
on = ts[other[curo]][0] if curo < len(other) else 9999999999
cur += min(an, bn, on)
if an <= bn and an <= on:
cura += 1
elif bn <= an and bn <= on:
curb += 1
else:
curo += 1
res, a, b, o = cur, cura, curb, curo
for i in range(x + 1, len(allL) + 1):
cur += ts[allL[i - 1]][0]
if cura > 0:
cura -= 1
cur -= ts[aliceL[cura]][0]
if curb > 0:
curb -= 1
cur -= ts[bobL[curb]][0]
if curo > 0:
curo -= 1
cur -= ts[other[curo]][0]
while cura + i + curb + curo < m:
an = ts[aliceL[cura]][0] if cura < len(aliceL) else 9999999999
bn = ts[bobL[curb]][0] if curb < len(bobL) else 9999999999
on = ts[other[curo]][0] if curo < len(other) else 9999999999
cur += min(an, bn, on)
if an <= bn and an <= on:
cura += 1
elif bn <= an and bn <= on:
curb += 1
else:
curo += 1
if res > cur:
res, x, a, b, o = cur, i, cura, curb, curo
print(res)
for i in range(x):
print(allL[i], end=" ")
for i in range(a):
print(aliceL[i], end=" ")
for i in range(b):
print(bobL[i], end=" ")
for i in range(o):
print(other[i], end=" ") | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING ASSIGN VAR VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER LIST LIST LIST LIST ASSIGN VAR LIST VAR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP VAR NUMBER LIST VAR VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP NUMBER BIN_OP VAR FUNC_CALL VAR VAR BIN_OP VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR VAR NUMBER BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR VAR WHILE BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER NUMBER VAR FUNC_CALL VAR VAR VAR VAR IF VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER NUMBER IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER WHILE BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER NUMBER VAR FUNC_CALL VAR VAR VAR VAR IF VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR STRING FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR STRING FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR STRING FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR 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 | def func(n, k):
common = []
bob = []
alice = []
for i in range(n):
t, a, b = map(int, input().split())
if a and b:
common.append(t)
elif a:
alice.append(t)
elif b:
bob.append(t)
if len(common) + len(alice) < k or len(common) + len(bob) < k:
print(-1)
return
common.sort()
alice.sort()
bob.sort()
c_num = min(k, len(common))
num = k - c_num
while (
c_num != 0
and num < len(alice)
and num < len(bob)
and common[c_num - 1] > alice[num] + bob[num]
):
c_num -= 1
num += 1
print(sum(common[:c_num]) + sum(alice[:num]) + sum(bob[:num]))
for i in range(1):
n, k = map(int, input().split())
func(n, k) | FUNC_DEF 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 RETURN 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 WHILE VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR 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 FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR 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 = input().split(" ")
n, k = int(n), int(k)
S = []
for i in range(n):
S.append(input().split(" "))
A = []
B = []
C = []
for i in S:
if i[1] == "1" and i[2] == "1":
C.append(int(i[0]))
elif i[1] == "1" and i[2] == "0":
A.append(int(i[0]))
elif i[1] == "0" and i[2] == "1":
B.append(int(i[0]))
m = min(len(A), len(B))
if len(C) + m < k:
print(-1)
else:
A.sort()
B.sort()
for i in range(m):
C.append(A[i] + B[i])
ans = 0
C.sort()
for i in range(k):
ans += C[i]
print(ans) | ASSIGN VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR IF VAR NUMBER STRING VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER STRING VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER STRING VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR ASSIGN 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 | n, k = [int(x) for x in input().split()]
a, b, ab = [0], [0], [0]
for _ in range(n):
p, q, r = [int(x) for x in input().split()]
if q == 1:
if r == 1:
ab.append(p)
else:
a.append(p)
elif r == 1:
b.append(p)
if len(ab) - 1 + min(len(a) - 1, len(b) - 1) < k:
print(-1)
exit()
a.sort()
b.sort()
ab.sort()
sumo = 0
for i in range(len(a)):
sumo += a[i]
a[i] = sumo
sumo = 0
for i in range(len(b)):
sumo += b[i]
b[i] = sumo
sumo = 0
for i in range(len(ab)):
sumo += ab[i]
ab[i] = sumo
minans = int(10000000000.0)
for kp in range(max(0, k + 1 - min(len(a), len(b))), min(len(ab), k + 1)):
minans = min(minans, a[k - kp] + b[k - kp] + ab[kp])
print(minans) | ASSIGN VAR VAR FUNC_CALL VAR 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 VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER 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 FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR VAR BIN_OP 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
n, k = map(int, stdin.readline().split())
ba, bb, bc = [], [], []
for _ in range(n):
t, a, b = map(int, stdin.readline().split())
if a and b:
bc.append(t)
elif a:
ba.append(t)
elif b:
bb.append(t)
ba.sort()
bb.sort()
bc.sort()
a, b, c = 0, 0, 0
T = 0
while k:
if c < len(bc) and a < len(ba) and b < len(bb):
if bc[c] < ba[a] + bb[b]:
T += bc[c]
c += 1
k -= 1
else:
T += ba[a] + bb[b]
a += 1
b += 1
k -= 1
elif c < len(bc):
T += bc[c]
c += 1
k -= 1
elif a < len(ba) and b < len(bb):
T += ba[a] + bb[b]
a += 1
b += 1
k -= 1
else:
T = -1
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 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 VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER 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 | n, k = [int(x) for x in input().split()]
books = []
books_a = []
books_b = []
for i in range(n):
b = [int(x) for x in input().split()]
if b[1] == 1 and b[2] == 1:
books.append(b[0])
elif b[1] == 1:
books_a.append(b[0])
elif b[2] == 1:
books_b.append(b[0])
books_a = sorted(books_a)
books_b = sorted(books_b)
z = min(len(books_a), len(books_b))
for j in range(z):
books.append(books_a[j] + books_b[j])
books = sorted(books)
ans = 0
if len(books) < k:
print(-1)
else:
for f in range(k):
ans += books[f]
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 FUNC_CALL VAR 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 ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR 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 ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL 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
inp = [int(x) for x in sys.stdin.read().split()]
ii = 0
n = inp[ii]
ii += 1
k = inp[ii]
ii += 1
T = inp[ii + 0 : ii + 3 * n : 3]
A = inp[ii + 1 : ii + 3 * n : 3]
B = inp[ii + 2 : ii + 3 * n : 3]
TA = []
TB = []
TAB = []
for i in range(n):
if A[i] and B[i]:
TAB.append(T[i])
elif A[i]:
TA.append(T[i])
elif B[i]:
TB.append(T[i])
def cumsummer(A):
cumsum = [0]
for a in A:
cumsum.append(cumsum[-1] + a)
return cumsum
TA.sort()
TB.sort()
TAB.sort()
n = len(TA)
m = len(TB)
nm = len(TAB)
TA = cumsummer(TA)
TB = cumsummer(TB)
TAB = cumsummer(TAB)
time = inf = 10**9 * 2 + 100
for x in range(min(k, nm) + 1):
y = z = k - x
if n < y or m < z:
continue
time = min(time, TAB[x] + TA[y] + TB[z])
print(time if time != inf else -1) | IMPORT ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP NUMBER VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP NUMBER VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP NUMBER VAR NUMBER 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 VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR 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 ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP BIN_OP BIN_OP NUMBER NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR 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 abc(l, n, i, x, y, m, k):
if x >= k and y >= k:
return m
if i == n:
return -1
ans = -1
ans = abc(l, n, i + 1, x, y, m, k)
c = abc(l, n, i + 1, x + l[i][1], y + l[i][2], m + l[i][0], k)
if c != -1:
if ans == -1:
ans = c
else:
ans = min(ans, c)
return ans
n, k = map(int, input().split())
l = []
for i in range(n):
l.append(list(map(int, input().split())))
d = {}
d[0, 0] = []
d[0, 1] = []
d[1, 0] = []
d[1, 1] = []
for i in l:
d[i[1], i[2]] += [i[0]]
for i in d:
d[i] = sorted(d[i], reverse=True)
ans = 0
while k > 0:
if len(d[1, 1]) == 0:
if len(d[0, 1]) == 0 or len(d[1, 0]) == 0:
break
ans += d[0, 1][-1] + d[1, 0][-1]
del d[0, 1][-1]
del d[1, 0][-1]
k -= 1
elif len(d[0, 1]) == 0 or len(d[1, 0]) == 0:
ans += d[1, 1][-1]
del d[1, 1][-1]
k -= 1
else:
x = d[0, 1][-1] + d[1, 0][-1]
if x <= d[1, 1][-1]:
ans += d[0, 1][-1] + d[1, 0][-1]
del d[0, 1][-1]
del d[1, 0][-1]
k -= 1
else:
ans += d[1, 1][-1]
del d[1, 1][-1]
k -= 1
if k == 0:
print(ans)
else:
print(-1) | FUNC_DEF IF VAR VAR VAR VAR RETURN VAR IF VAR VAR RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR NUMBER BIN_OP VAR VAR VAR NUMBER BIN_OP VAR VAR VAR NUMBER VAR IF VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR 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 DICT ASSIGN VAR NUMBER NUMBER LIST ASSIGN VAR NUMBER NUMBER LIST ASSIGN VAR NUMBER NUMBER LIST ASSIGN VAR NUMBER NUMBER LIST FOR VAR VAR VAR VAR NUMBER VAR NUMBER LIST VAR NUMBER FOR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER IF FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER NUMBER NUMBER FUNC_CALL VAR VAR NUMBER NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR NUMBER NUMBER NUMBER VAR NUMBER NUMBER NUMBER VAR NUMBER NUMBER NUMBER VAR NUMBER IF FUNC_CALL VAR VAR NUMBER NUMBER NUMBER FUNC_CALL VAR VAR NUMBER NUMBER NUMBER VAR VAR NUMBER NUMBER NUMBER VAR NUMBER NUMBER NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR NUMBER NUMBER NUMBER VAR NUMBER NUMBER NUMBER VAR NUMBER NUMBER NUMBER VAR NUMBER VAR VAR NUMBER NUMBER NUMBER VAR NUMBER NUMBER NUMBER 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 | def mi():
return map(int, input().split())
n, k = mi()
a, b, ab = [], [], []
for i in range(n):
t, al, bl = mi()
lab, la, lb = 0, 0, 0
if al and bl:
ab.append(t)
elif al:
a.append(t)
elif bl:
b.append(t)
b.sort(), a.sort(), ab.sort()
la, lb, lab = len(a), len(b), len(ab)
for i in range(1, lb):
b[i] += b[i - 1]
for i in range(1, la):
a[i] += a[i - 1]
ans = 10000000000.0
if len(a) >= k and len(b) >= k:
ans = a[k - 1] + b[k - 1]
for i in range(lab):
if i > 0:
ab[i] += ab[i - 1]
if not k - 1 - i:
ans = min(ans, ab[i])
if k - 1 - i > 0 and min(len(a), len(b)) >= k - 1 - i:
ans = min(ans, ab[i] + a[k - i - 2] + b[k - i - 2])
if ans == 10000000000.0:
ans = -1
print(ans) | FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER 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 FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF BIN_OP BIN_OP VAR NUMBER VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR 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 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 | book, k = map(int, input().split())
both, bob, alice = dict(), dict(), dict()
for i in range(book):
time, x, y = map(int, input().split())
if x == 1 and y == 0:
alice[i] = time
elif x == 0 and y == 1:
bob[i] = time
elif x == 1 and y == 1:
both[i] = time
if len(both) + len(alice) < k or len(both) + len(bob) < k:
print(-1)
exit()
else:
alice = sorted(alice.items(), key=lambda x: x[1])
bob = sorted(bob.items(), key=lambda x: x[1])
both = sorted(both.items(), key=lambda x: x[1])
count, x, y, z, time = 0, 0, 0, 0, 0
while count < k:
if x < len(alice) and y < len(bob) and z < len(both):
if alice[x][1] + bob[y][1] < both[z][1]:
time += alice[x][1] + bob[y][1]
x += 1
y += 1
count += 1
else:
time += both[z][1]
count += 1
z += 1
elif z >= len(both):
time += alice[x][1] + bob[y][1]
x += 1
y += 1
count += 1
else:
time += both[z][1]
count += 1
z += 1
print(time) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR 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 IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR 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 ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER NUMBER NUMBER NUMBER NUMBER WHILE VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF 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 NUMBER VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER 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 | import sys
input = sys.stdin.readline
n, k = map(int, input().split())
both = []
ali = []
bli = []
for i in range(n):
t, a, b = map(int, input().split())
if a == 1 and b == 1:
both.append(t)
if a == 1 and b == 0:
ali.append(t)
if a == 0 and b == 1:
bli.append(t)
lboth = len(both)
lali = len(ali)
lbli = len(bli)
if lboth + lali < k or lboth + lbli < k:
print(-1)
else:
both.sort(reverse=True)
ali.sort(reverse=True)
bli.sort(reverse=True)
ans = 0
for i in range(k):
if lboth != 0 and lali != 0 and lbli != 0:
if both[-1] <= ali[-1] + bli[-1]:
ans += both[-1]
both.pop()
lboth -= 1
else:
ans += ali[-1] + bli[-1]
lali -= 1
lbli -= 1
ali.pop()
bli.pop()
elif lboth == 0:
ans += ali[-1] + bli[-1]
lali -= 1
lbli -= 1
ali.pop()
bli.pop()
elif lali == 0 or lbli == 0:
ans += both[-1]
both.pop()
lboth -= 1
print(ans) | IMPORT 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 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 NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR NUMBER VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL 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())
x, y, z = [], [], []
for _ in range(n):
t, a, b = map(int, input().split())
if a == b == 1:
x.append(t)
elif a == 1:
y.append(t)
elif b == 1:
z.append(t)
lx, ly, lz = len(x), len(y), len(z)
x.sort()
y.sort()
z.sort()
u, v = len(x), min(len(y), len(z))
ans = 0
if u + v >= k:
i, j = 0, 0
for _ in range(k):
if i < u and j < v:
if x[i] <= y[j] + z[j]:
ans += x[i]
i += 1
else:
ans += y[j] + z[j]
j += 1
elif i < u:
ans += x[i]
i += 1
elif j < v:
ans += y[j] + z[j]
j += 1
else:
ans = -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 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 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 ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR VAR VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR 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 IF VAR VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR VAR VAR 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 | n, k = map(int, input().split())
ab, a, b = [], [], []
for _ in range(n):
t, aa, bb = map(int, input().split())
if aa and bb:
ab.append(t)
elif aa:
a.append(t)
elif bb:
b.append(t)
a.sort()
b.sort()
zz = min(len(a), len(b))
for i in range(zz):
ab.append(a[i] + b[i])
print(-1 if len(ab) < k else sum(sorted(ab)[:k])) | 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 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 FUNC_CALL VAR VAR VAR NUMBER 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
class E1ReadingBooksEasyVersion:
def solve(self, tc=0):
n, k = [int(_) for _ in input().split()]
books = []
for i in range(n):
books.append([int(_) for _ in input().split()])
a = []
b = []
both = []
for i in range(n):
book = books[i]
if book[1] and book[2]:
both.append(book[0])
elif book[1]:
a.append(book[0])
elif book[2]:
b.append(book[0])
a.sort()
b.sort()
both.sort()
prea = [0] * (len(a) + 1)
for i in range(1, len(a) + 1):
prea[i] = prea[i - 1] + a[i - 1]
preb = [0] * (len(b) + 1)
for i in range(1, len(b) + 1):
preb[i] = preb[i - 1] + b[i - 1]
preboth = [0] * (len(both) + 1)
for i in range(1, len(both) + 1):
preboth[i] = preboth[i - 1] + both[i - 1]
ans = float("inf")
for i in range(len(both) + 1):
if i > k:
break
if len(a) >= k - i and len(b) >= k - i:
ans = min(ans, preboth[i] + prea[k - i] + preb[k - i])
print(ans if ans != float("inf") else -1)
solver = E1ReadingBooksEasyVersion()
input = sys.stdin.readline
solver.solve() | IMPORT CLASS_DEF FUNC_DEF NUMBER 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 FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER 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 FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP 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 BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP 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 BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR IF FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP 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 FUNC_CALL VAR STRING VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN 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 = map(int, input().split())
ali = list()
bob = list()
both = list()
for i in range(n):
l = list(map(int, input().split()))
if l[1] == 1 and l[2] == 1:
both.append(l[0])
elif l[1] == 1:
ali.append(l[0])
elif l[2] == 1:
bob.append(l[0])
ali.sort(reverse=True)
bob.sort(reverse=True)
both.sort(reverse=True)
a = len(ali)
b = len(bob)
if a > b:
a = b
bt = len(both)
if a + bt < k:
print(-1)
else:
count = 0
i, j = -1, -1
a, bt = -a, -bt
while i >= a and j >= bt and k > 0:
if ali[i] + bob[i] < both[j]:
count += ali[i] + bob[i]
k -= 1
i -= 1
else:
count += both[j]
j -= 1
k -= 1
if j < bt:
while i >= a and k > 0:
count += ali[i] + bob[i]
k -= 1
i -= 1
else:
while j >= bt and k > 0:
count += both[j]
j -= 1
k -= 1
print(count) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR 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 EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR WHILE VAR VAR VAR VAR 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 IF VAR VAR WHILE VAR VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER WHILE VAR VAR 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 | a, k = map(int, input().split())
l1 = []
l2 = []
l3 = []
for i in range(a):
b, c, d = map(int, input().split())
if c == 1 and d == 1:
l3.append(b)
elif c == 1:
l1.append(b)
elif d == 1:
l2.append(b)
l1 = sorted(l1)
l2 = sorted(l2)
l3 = sorted(l3)
if len(l1) + len(l3) < k or len(l2) + len(l3) < k:
print(-1)
else:
d = 0
k1 = 0
k2 = 0
k3 = 0
for i in range(k):
if len(l3) > k3 and (len(l1) > k1 and len(l2) > k2):
if l3[k3] < l1[k1] + l2[k2]:
d += l3[k3]
k3 += 1
else:
d += l1[k1] + l2[k2]
k1 += 1
k2 += 1
elif k3 < len(l3):
d += l3[k3]
k3 += 1
else:
d += l1[k1] + l2[k2]
k1 += 1
k2 += 1
print(d) | 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 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 ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR 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 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 | import sys
def main():
import sys
input = sys.stdin.buffer.readline
n, k = map(int, input().split())
tab = [tuple(map(int, input().split())) for i in range(n)]
tab.sort(reverse=True)
A = [tab[i][0] for i in range(n) if tab[i][1] == 1 and tab[i][2] == 0]
B = [tab[i][0] for i in range(n) if tab[i][1] == 0 and tab[i][2] == 1]
AB = [tab[i][0] for i in range(n) if tab[i][1] == tab[i][2] == 1]
count = 0
ans = 0
ab = 0
apb = 0
while count < k:
if len(AB) > 0:
ab = AB[-1]
else:
ab = 10**10
if len(A) > 0 and len(B) > 0:
apb = A[-1] + B[-1]
else:
apb = 10**10
if ab < apb:
ans += ab
if len(AB):
AB.pop()
else:
ans += apb
if len(A):
A.pop()
if len(B):
B.pop()
count += 1
if ans >= 10**10:
print(-1)
else:
print(ans)
main() | IMPORT FUNC_DEF 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 NUMBER ASSIGN VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER IF VAR VAR VAR VAR IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP NUMBER NUMBER EXPR FUNC_CALL 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 | n, k = map(int, input().split(" "))
both = []
bob = []
alice = []
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)
if len(both) != 0:
both.sort()
if len(alice) != 0:
alice.sort()
if len(bob) != 0:
bob.sort()
if len(both) + len(alice) < k or len(both) + len(bob) < k:
print(-1)
else:
i = 0
j = 0
ans = 0
q = 0
w = 0
talice = 0
tbob = 0
while i < len(alice) and j < len(bob) and q < k and w < len(both):
if alice[i] + bob[j] <= both[w]:
ans += alice[i] + bob[j]
talice += 1
tbob += 1
i += 1
j += 1
else:
ans += both[w]
talice += 1
tbob += 1
w += 1
q += 1
z = talice
if i == len(alice) or j == len(bob):
ans += sum(both[w : w + k - z])
elif w == len(both):
ans += sum(alice[i : i + k - z])
ans += sum(bob[j : j + k - z])
print(ans) | ASSIGN VAR VAR FUNC_CALL 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 FUNC_CALL FUNC_CALL VAR STRING 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 VAR NUMBER EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER 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 ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR 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 VAR NUMBER ASSIGN VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR IF VAR FUNC_CALL 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())
x = []
y = []
z = []
for _ in range(n):
t, a, b = map(int, input().split())
if a == 1 and b == 1:
x.append(t)
elif a == 1 and b == 0:
y.append(t)
elif a == 0 and b == 1:
z.append(t)
x.sort()
y.sort()
z.sort()
kk = []
p = min(len(y), len(z))
for i in range(p):
kk.append(y[i] + z[i])
if len(kk) + len(x) < k:
print(-1)
else:
i = 0
j = 0
c = []
while True:
if i == len(kk):
c.extend(x[j:])
break
if j == len(x):
c.extend(kk[i:])
break
if kk[i] > x[j]:
c.append(x[j])
j += 1
elif kk[i] == x[j]:
c.append(x[j])
c.append(x[j])
i += 1
j += 1
else:
c.append(kk[i])
i += 1
print(sum(c[:k])) | 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 ASSIGN VAR LIST 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 BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST WHILE NUMBER IF VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER 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 | n, k = map(int, input().split())
main = []
for i in range(n):
l = list(map(int, input().split()))
main.append(l)
list01 = []
list10 = []
list11 = []
for i in main:
if i[1] == 0 and i[2] == 1:
list01.append(i)
elif i[1] == 1 and i[2] == 0:
list10.append(i)
elif i[1] == 1 and i[2] == 1:
list11.append(i)
list01.sort()
list10.sort()
newlist11 = []
l1 = len(list01)
l2 = len(list10)
sum = 0
for i in range(0, min(l1, l2)):
sum = list01[i][0] + list10[i][0]
newlist11.append([sum, 1, 1])
finallist = []
finallist = list11 + newlist11
finallist.sort()
add = 0
if len(finallist) < k:
print(-1)
else:
for i in range(0, k):
add += finallist[i][0]
print(add) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR 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 IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR LIST VAR NUMBER NUMBER ASSIGN VAR LIST ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL 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 = list(map(int, input().split()))
books = [[], [], [], []]
for i in range(n):
s = list(map(int, input().split()))
if s[1] == 1 and s[2] == 1:
books[2].append(s[0])
elif s[2] == 1:
books[1].append(s[0])
elif s[1] == 1:
books[0].append(s[0])
if len(books[0]) + len(books[2]) < k or len(books[1]) + len(books[2]) < k:
print(-1)
else:
for x in books:
x.sort()
t = 0
al = 0
bl = 0
fi = 0
si = 0
can = True
while al < k or bl < k:
if fi < len(books[0]) and fi < len(books[1]) and si < len(books[2]):
c1 = books[0][fi] + books[1][fi]
c2 = books[2][si]
if c1 < c2:
al += 1
bl += 1
t = t + c1
fi = fi + 1
else:
al += 1
bl += 1
t = t + c2
si = si + 1
elif fi == len(books[0]):
needed = k - al
t = t + sum(books[2][si : si + needed])
break
elif fi == len(books[1]):
needed = k - bl
t = t + sum(books[2][si : si + needed])
break
elif si == len(books[2]):
needed = k - al
t = t + sum(books[0][fi : fi + needed])
needed = k - bl
t = t + sum(books[1][fi : fi + needed])
break
else:
can = False
break
if can:
print(t)
else:
print(-1) | ASSIGN VAR VAR FUNC_CALL 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 FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER VAR NUMBER IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER VAR NUMBER IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER FOR VAR 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 IF VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR VAR NUMBER VAR ASSIGN VAR VAR NUMBER VAR IF VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR VAR IF VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR VAR IF VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR VAR 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 = []
for _ in range(n):
x = list(map(int, input().split()))
if x[2] == 1 and x[1] == 1:
both.append(x[0])
elif x[1]:
alice.append(x)
elif x[2]:
bob.append(x)
alice.sort(key=lambda x: x[0])
bob.sort(key=lambda x: x[0])
both.sort()
tgt = []
for i in range(min(len(alice), len(bob))):
tgt.append(alice[i][0] + bob[i][0])
p1, p2 = 0, 0
count = 0
ans = 0
if len(tgt) + len(both) < k:
print(-1)
else:
while count < k:
if p2 == len(both) or p1 < len(tgt) and tgt[p1] <= both[p2]:
ans += tgt[p1]
p1 += 1
else:
ans += both[p2]
p2 += 1
count += 1
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR LIST 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 EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER 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 BIN_OP VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER WHILE VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR 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 | import sys
def get_ints():
return map(int, sys.stdin.readline().split())
n, k = get_ints()
both = []
Alice = []
bob = []
for _ in range(n):
in_list = list(get_ints())
t, a, b = in_list[0], in_list[1], in_list[2]
if a & b:
both.append(t)
elif a:
Alice.append(t)
elif b:
bob.append(t)
Alice.sort()
bob.sort()
l = min(len(Alice), len(bob))
for i in range(l):
both.append(Alice[i] + bob[i])
if len(both) < k:
print(-1)
else:
both.sort()
print(sum(both[:k])) | IMPORT FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR 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 ASSIGN VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF BIN_OP 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 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 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 | def solution():
aLike = []
bLike = []
bothLike = []
lines, k = input().strip().split()
lines = int(lines)
k = int(k)
for i in range(lines):
cost, isALike, isBLike = input().strip().split()
cost = int(cost)
isALike = isALike == "1"
isBLike = isBLike == "1"
if isALike and isBLike:
bothLike.append(cost)
elif isALike:
aLike.append(cost)
elif isBLike:
bLike.append(cost)
aLike.sort()
bLike.sort()
bothLike.sort()
aLike.reverse()
bLike.reverse()
bothLike.reverse()
cost = 0
while len(bothLike) > 0 and k > 0:
if len(aLike) > 0 and len(bLike) > 0:
if bothLike[-1] <= aLike[-1] + bLike[-1]:
cost += bothLike.pop()
k -= 1
else:
cost += aLike.pop()
cost += bLike.pop()
k -= 1
else:
cost += bothLike.pop()
k -= 1
while k > 0:
if len(aLike) > 0 and len(bLike) > 0:
cost += aLike.pop()
cost += bLike.pop()
k -= 1
else:
return -1
return cost
print(solution()) | FUNC_DEF ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR STRING ASSIGN VAR VAR STRING 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 EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER WHILE FUNC_CALL VAR VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER IF VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER WHILE VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER RETURN NUMBER RETURN 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 = [int(x) for x in input().split()]
bookA = []
bookB = []
bookC = []
for i in range(n):
a, b, c = [int(x) for x in input().split()]
if b == c == 1:
bookC.append((a, b, c))
if b == 1 != c:
bookA.append((a, b, c))
if c == 1 != b:
bookB.append((a, b, c))
bookA.sort(key=lambda x: x[0], reverse=True)
bookB.sort(key=lambda x: x[0], reverse=True)
bookC.sort(key=lambda x: x[0], reverse=True)
totalTime = 0
totalLikes = 0
while True:
if totalLikes >= k:
break
if len(bookA) == len(bookC) == len(bookB) == 0:
break
elif len(bookC) > 0 and (
len(bookA) == 0
or len(bookB) == 0
or bookC[len(bookC) - 1][0]
<= bookA[len(bookA) - 1][0] + bookB[len(bookB) - 1][0]
):
totalTime += bookC[len(bookC) - 1][0]
bookC.pop()
totalLikes += 1
elif len(bookA) > 0 and len(bookB) > 0:
totalTime += bookA[len(bookA) - 1][0] + bookB[len(bookB) - 1][0]
totalLikes += 1
bookA.pop()
bookB.pop()
else:
break
if totalLikes >= k:
print(totalTime)
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 NUMBER EXPR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE NUMBER IF VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR 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 | def solve(n, k, a, b, c):
a.sort()
b.sort()
c.sort()
iab = min(k, len(a), len(b))
ic = k - iab
if ic > len(c):
return -1
res = 0
for i in range(ic):
res += c[i]
for i in range(iab):
res += a[i] + b[i]
iab -= 1
while ic < len(c) and iab >= 0 and c[ic] < a[iab] + b[iab]:
res -= a[iab] + b[iab] - c[ic]
ic += 1
iab -= 1
return res
current_str = input().split(" ")
n = int(current_str[0])
k = int(current_str[1])
a = []
b = []
c = []
for i in range(n):
current_str = input().split(" ")
t = int(current_str[0])
aa = int(current_str[1])
bb = int(current_str[2])
if aa and bb:
c.append(t)
elif aa:
a.append(t)
elif bb:
b.append(t)
print(solve(n, k, a, b, c)) | FUNC_DEF EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR FUNC_CALL VAR VAR RETURN NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR BIN_OP VAR VAR VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER 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 FUNC_CALL VAR 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())
both = []
alice = []
bob = []
none = []
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)
else:
none.append(t)
flag = 0
both.sort()
alice.sort()
bob.sort()
if len(both) < k:
if len(alice) + len(both) < k or len(bob) + len(both) < k:
print(-1)
flag = 1
bothpt = 0
bobpt = 0
alicept = 0
count = 0
if flag == 0:
for i in range(k):
if alicept >= len(alice) or bobpt >= len(bob):
count += both[bothpt]
bothpt += 1
continue
if bothpt < len(both) and both[bothpt] <= alice[alicept] + bob[bobpt]:
count += both[bothpt]
bothpt += 1
else:
count += alice[alicept] + bob[bobpt]
alicept += 1
bobpt += 1
print(count) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST 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 VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF FUNC_CALL VAR 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 NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER IF 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 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()]
arr = []
for i in range(n):
arr.append([int(i) for i in input().split()])
one_one = []
zero_one = []
one_zero = []
for i in arr:
if i[2] == i[1] == 1:
one_one.append(i[0])
elif i[1] == 1:
one_zero.append(i[0])
elif i[2] == 1:
zero_one.append(i[0])
one_one.append(10**20)
one_zero.append(10**20)
zero_one.append(10**20)
one_one.sort()
zero_one.sort()
one_zero.sort()
ptr1 = ptr2 = ptr3 = 0
n1 = 0
ans = 0
while n1 < k:
if one_one[ptr1] < one_zero[ptr2] + zero_one[ptr3]:
ans += one_one[ptr1]
ptr1 += 1
n1 += 1
else:
ans += one_zero[ptr2] + zero_one[ptr3]
ptr2 += 1
ptr3 += 1
n1 += 1
if ans >= 10**20:
ans = -1
break
print(ans) | 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 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 EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR BIN_OP NUMBER 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 | n, k = map(int, input().split())
v1, v2, v12 = [], [], []
for _ in range(n):
t, l1, l2 = map(int, input().split())
if l1 == 1 and l2 == 1:
v12.append(t)
elif l1 == 1:
v1.append(t)
elif l2 == 1:
v2.append(t)
v1.sort()
v2.sort()
v12.sort()
x, y, z = 0, 0, 0
ans = 0
sb = k - min(len(v1), len(v2))
sb = sb if sb > 0 else 0
k -= sb
if sb > len(v12):
print(-1)
else:
for z in range(sb):
ans += v12[z]
z = sb
while k > 0 and z < len(v12):
if v12[z] <= v1[x] + v2[y]:
ans += v12[z]
z += 1
else:
ans += v1[x] + v2[y]
x += 1
y += 1
k -= 1
if k <= len(v1) - x and k <= len(v2) - y:
while k > 0:
ans += v1[x] + v2[y]
x += 1
y += 1
k -= 1
print(ans)
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 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 ASSIGN VAR BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER VAR NUMBER VAR VAR IF VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR WHILE VAR NUMBER 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 VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR WHILE VAR NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER 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 | l = list(map(int, input().split()))
n, y = l[0], l[1]
fir = []
sec = []
bot = []
for _ in range(n):
k = list(map(int, input().split()))
if k[1] == 1 and k[2] == 1:
bot.append(k[0])
elif k[1] == 1 and k[2] == 0:
fir.append(k[0])
elif k[1] == 0 and k[2] == 1:
sec.append(k[0])
f = len(fir)
s = len(sec)
b = len(bot)
if f + b < y or s + b < y:
print(-1)
else:
fir.sort()
sec.sort()
bot.sort()
if f < y:
fir += (y - f) * [10001]
else:
fir = fir[0:y]
if s < y:
sec += (y - s) * [10001]
else:
sec = sec[0:y]
val = sum(fir) + sum(sec)
fir.reverse()
sec.reverse()
for i in range(y):
try:
if fir[i] + sec[i] > bot[i]:
val = val - fir[i] - sec[i] + bot[i]
except:
break
print(val) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER VAR NUMBER 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 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 IF VAR VAR VAR BIN_OP BIN_OP VAR VAR LIST NUMBER ASSIGN VAR VAR NUMBER VAR IF VAR VAR VAR BIN_OP BIN_OP VAR VAR LIST NUMBER ASSIGN VAR VAR NUMBER VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR 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 | t = 1
while t > 0:
n, k = map(int, input().split())
a = []
b = []
c = []
ans = 0
for i in range(n):
ti, al, bo = map(int, input().split())
if al == 1 and bo == 1:
c.append(ti)
elif al == 1:
a.append(ti)
elif bo == 1:
b.append(ti)
a.sort()
b.sort()
for i in range(min(len(a), len(b))):
c.append(a[i] + b[i])
c.sort()
if len(c) < k:
print(-1)
else:
print(sum(c[:k]))
t -= 1 | ASSIGN VAR NUMBER WHILE VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR 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 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 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 IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR 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 | a = input().split(" ")
list1 = []
list2 = []
list3 = []
for i in range(int(a[0])):
b = input().split(" ")
if b[1] == "1":
if b[2] == "1":
list1.append(int(b[0]))
else:
list2.append(int(b[0]))
elif b[2] == "1":
list3.append(int(b[0]))
if len(list1) + min(len(list2), len(list3)) < int(a[1]):
print(-1)
else:
ans = 0
ans1 = 0
ans2 = 0
list1_index = 0
list2_index = 0
list3_index = 0
list1_len = len(list1)
list2_len = len(list2)
list3_len = len(list3)
list1.sort()
list2.sort()
list3.sort()
while ans < int(a[1]) or ans1 < int(a[1]):
if ans < int(a[1]) and ans1 < int(a[1]) and list1_index < list1_len:
if list2_len <= list2_index or list3_len <= list3_index:
tag = int(a[1]) - min(ans, ans1)
ans2 += sum(list1[list1_index : list1_index + tag])
ans = int(a[1])
ans1 = int(a[1])
elif list1[list1_index] > list2[list2_index] + list3[list3_index]:
ans1 += 1
ans += 1
ans2 += list2[list2_index] + list3[list3_index]
list2_index += 1
list3_index += 1
else:
ans1 += 1
ans += 1
ans2 += list1[list1_index]
list1_index += 1
elif ans < int(a[1]):
if list2_len <= list2_index:
ans += 1
ans2 += list1[list1_index]
list1_index += 1
else:
ans += 1
ans2 += list2[list2_index]
list2_index += 1
elif ans1 < int(a[1]):
if list3_len <= list3_index:
ans1 += 1
ans2 += list1[list1_index]
list1_index += 1
else:
ans1 += 1
ans2 += list3[list3_index]
list3_index += 1
print(ans2) | ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING IF VAR NUMBER STRING IF VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER 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 EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR WHILE VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR VAR IF VAR VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR NUMBER 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 | n, k = list(map(int, input().split()))
a = []
b = []
ab = []
for _ in range(0, n):
t, x, y = list(map(int, input().split()))
if x == 1 and y == 1:
ab.append(t)
elif x == 1:
a.append(t)
elif y == 1:
b.append(t)
ab.sort(reverse=True)
b.sort(reverse=True)
a.sort(reverse=True)
ca = 0
cb = 0
t = 0
c = 0
u = 0
while len(a) != 0 and len(b) != 0:
if len(ab) > 0 and ab[-1] <= a[-1] + b[-1]:
t += ab.pop()
c += 1
else:
t = t + a.pop() + b.pop()
c += 1
if c == k:
u = 1
break
if u == 1:
print(t)
elif c + len(ab) < k:
print(-1)
else:
t += sum(ab[-(k - c) :])
print(t) | 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 NUMBER 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 NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER VAR FUNC_CALL 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 = map(int, input().split())
U = []
A = []
B = []
for i in range(n):
a, b, c = map(int, input().split())
if b == 1 and c == 1:
U.append(a)
elif b == 1 and c == 0:
A.append(a)
elif b == 0 and c == 1:
B.append(a)
A.sort()
B.sort()
U.sort()
for i in range(1, len(U)):
U[i] += U[i - 1]
for i in range(1, len(A)):
A[i] += A[i - 1]
for i in range(1, len(B)):
B[i] += B[i - 1]
f_ans = 10000000000.0
cnt = 0
ans = 0
for i in range(-1, len(U)):
cnt = i + 1
ans = 0
if i >= 0:
ans = U[i]
if k - cnt == 0:
f_ans = min(ans, f_ans)
break
if k - cnt > len(A) or k - cnt > len(B):
continue
ans += A[k - cnt - 1] + B[k - cnt - 1]
cnt = k
f_ans = min(f_ans, ans)
f_ans = min(ans, f_ans)
if k - cnt > 0:
print("-1")
else:
print(f_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 NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR IF BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR STRING 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())
l1 = []
l2 = []
l3 = []
for i in range(n):
t, a, b = map(int, input().split())
if (a, b) == (1, 1):
l3.append(t)
elif (a, b) == (0, 1):
l1.append(t)
elif (a, b) == (1, 0):
l2.append(t)
s = 0
l1.sort()
l2.sort()
l3.sort()
if len(l3) >= k:
s = sum(l3[:k])
p = 0
p1 = k - 1
else:
s = sum(l3)
v = k - len(l3)
if len(l1) >= v and len(l2) >= v:
s += sum(l1[:v]) + sum(l2[:v])
else:
print(-1)
exit()
p = v
p1 = len(l3) - 1
while p < len(l1) and p < len(l2) and p1 >= 0:
if l1[p] + l2[p] <= l3[p1]:
s = s - l3[p1] + l1[p] + l2[p]
p1 -= 1
p += 1
else:
break
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 VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER IF BIN_OP VAR VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR 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 = map(int, input().split())
x = list()
y = list()
z = list()
for _ in range(n):
t, a, b = map(int, input().split())
if a == 0 and b == 0:
continue
elif a == 1 and b == 0:
x.append(t)
elif a == 0 and b == 1:
y.append(t)
else:
z.append(t)
x.sort()
y.sort()
for i in range(min(len(x), len(y))):
z.append(x[i] + y[i])
z.sort()
if len(z) < k:
print("-1")
else:
res = 0
for i in range(0, k):
res += z[i]
print(res) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR 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 IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR 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 IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER 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())
a = []
b = []
ab = []
for i in " " * n:
t, x, y = map(int, input().split())
if x & y:
ab += [t]
elif x:
a += [t]
elif y:
b += [t]
for i, j in zip(sorted(a), sorted(b)):
ab += [i + j]
if len(ab) < k:
print(-1)
else:
print(sum(sorted(ab)[:k])) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR BIN_OP STRING VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP VAR VAR VAR LIST VAR IF VAR VAR LIST VAR IF VAR VAR LIST VAR FOR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR LIST BIN_OP 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 | n, k = map(int, input().split())
a = []
b = []
c = []
for i in range(n):
x, y, z = map(int, input().split())
if y == 1 and z != 1:
a.append(x)
elif y != 1 and z == 1:
b.append(x)
elif y == 1 and z == 1:
c.append(x)
else:
pass
if len(a) + len(c) < k or len(c) + len(b) < k:
ans = -1
else:
a.sort()
b.sort()
c.sort()
pt1, pt2, pt3 = 0, 0, 0
move1 = 0
move2 = 0
ans = 0
while True:
if len(c) == 0:
for i in range(k):
ans += a[i]
for i in range(k):
ans += b[i]
break
elif move1 == k and move2 == k:
break
elif move1 < k and move2 == k:
if pt1 < len(a) and pt3 < len(c):
if a[pt1] < c[pt3]:
ans += a[pt1]
pt1 += 1
else:
ans += c[pt3]
pt3 += 1
elif not pt1 < len(a) and pt3 < len(c):
ans += c[pt3]
pt3 += 1
else:
ans += a[pt1]
pt1 += 1
move1 += 1
elif move1 == k and move2 < k:
if pt2 < len(b) and pt3 < len(c):
if b[pt2] < c[pt3]:
ans += b[pt2]
pt2 += 1
else:
ans += c[pt3]
pt3 += 1
elif not pt2 < len(b) and pt3 < len(c):
ans += c[pt3]
pt3 += 1
else:
ans += b[pt2]
pt2 += 1
move2 += 1
elif pt1 < len(a) and pt2 < len(b) and pt3 < len(c):
if a[pt1] + b[pt2] < c[pt3]:
ans += a[pt1]
move1 += 1
ans += b[pt2]
move2 += 1
pt1 += 1
pt2 += 1
else:
ans += c[pt3]
move1 += 1
move2 += 1
pt3 += 1
elif pt3 < len(c):
ans += c[pt3]
pt3 += 1
move1 += 1
move2 += 1
else:
if pt1 < len(a):
ans += a[pt1]
pt1 += 1
move1 += 1
if pt2 < len(b):
ans += b[pt2]
pt2 += 1
move2 += 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 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 ASSIGN VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE NUMBER IF FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR VAR VAR VAR IF VAR VAR VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER IF 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 VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR FUNC_CALL 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 = [int(_) for _ in input().split()]
alice = []
bob = []
both = []
for i in range(n):
book = [int(_) for _ in input().split()]
if book[1] == book[2] == 1:
both.append(book[0])
elif book[1] == 1:
alice.append(book[0])
elif book[2] == 1:
bob.append(book[0])
alice.sort()
bob.sort()
both.sort()
a = b = c = t = 0
for i in range(k):
if (a == len(alice) or b == len(bob)) and c == len(both):
t = -1
break
if c < len(both) and (
a == len(alice) or b == len(bob) or both[c] < alice[a] + bob[b]
):
t += both[c]
c += 1
else:
t += alice[a] + bob[b]
a += 1
b += 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 FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR 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 EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR 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 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 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
inp = [int(x) for x in sys.stdin.read().split()]
ii = 0
seg = [0] * 200000
def offset(x):
return x + 100000
def encode(x, y):
return x * 200002 + y
def decode(x):
return x // 200002, x % 200002
def upd(node, L, R, pos, val):
while L < R:
seg[node] += val
seg[offset(node)] += val * pos
if L + 1 == R:
break
M = (L + R) // 2
node <<= 1
if pos < M:
R = M
else:
L = M
node += 1
def query(node, L, R, k):
ret = 0
while L < R:
if k == 0:
return ret
if seg[node] == k:
return ret + seg[offset(node)]
if L + 1 == R:
return ret + k * L
M = (L + R) // 2
node <<= 1
if seg[node] >= k:
R = M
else:
ret += seg[offset(node)]
k -= seg[node]
L = M
node += 1
return ret
n, m, k = inp[ii : ii + 3]
ii += 3
A, B, both, neither = [], [], [], []
for i in range(n):
t, a, b = inp[ii : ii + 3]
ii += 3
if a == 0 and b == 0:
neither.append(encode(t, i + 1))
if a == 1 and b == 0:
A.append(encode(t, i + 1))
if a == 0 and b == 1:
B.append(encode(t, i + 1))
if a == 1 and b == 1:
both.append(encode(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] // 200002
upd(1, 0, 10001, both[i] // 200002, -1)
for i in range(p2):
sum += A[i] // 200002 + B[i] // 200002
upd(1, 0, 10001, A[i] // 200002, -1)
upd(1, 0, 10001, B[i] // 200002, -1)
ans = query(1, 0, 10001, m - 2 * k + p1) + sum
while p1 > 0:
if p2 == min(len(A), len(B)):
break
upd(1, 0, 10001, A[p2] // 200002, -1)
sum += A[p2] // 200002
upd(1, 0, 10001, B[p2] // 200002, -1)
sum += B[p2] // 200002
upd(1, 0, 10001, both[p1 - 1] // 200002, 1)
sum -= both[p1 - 1] // 200002
p2 += 1
p1 -= 1
if m - 2 * k + p1 < 0:
break
Q = query(1, 0, 10001, m - 2 * k + p1)
if ans > sum + Q:
ans = sum + Q
ch = p1
print(ans)
ind = (
[(both[i] % 200002) for i in range(ch)]
+ [(A[i] % 200002) for i in range(k - ch)]
+ [(B[i] % 200002) 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] % 200002) for i in range(m - 2 * k + ch)]
print(" ".join(str(x) for x in ind)) | IMPORT ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER FUNC_DEF RETURN BIN_OP VAR NUMBER FUNC_DEF RETURN BIN_OP BIN_OP VAR NUMBER VAR FUNC_DEF RETURN BIN_OP VAR NUMBER BIN_OP VAR NUMBER FUNC_DEF WHILE VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER WHILE VAR VAR IF VAR NUMBER RETURN VAR IF VAR VAR VAR RETURN BIN_OP VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER VAR RETURN BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER RETURN VAR ASSIGN VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR LIST LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR 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 BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER BIN_OP VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER BIN_OP VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR NUMBER NUMBER NUMBER BIN_OP BIN_OP VAR BIN_OP NUMBER VAR VAR VAR WHILE VAR NUMBER IF VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER BIN_OP VAR VAR NUMBER NUMBER VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER BIN_OP VAR VAR NUMBER NUMBER VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR BIN_OP 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 ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP 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 BIN_OP 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 | import sys
s = sys.stdin.readline().split()
n, m, k = int(s[0]), int(s[1]), int(s[2])
all = []
All = []
Alice = []
Bob = []
Both = []
none = []
z = 1
while n:
i = sys.stdin.readline().split()
x = 3
i.append(z)
while x:
i[x - 1] = int(i[x - 1])
x -= 1
all.append(i)
if i[1] == i[2]:
if i[1] == 0:
none.append(i)
else:
Both.append(i)
elif i[1] == 0:
Bob.append(i)
else:
Alice.append(i)
z += 1
n -= 1
Alice.sort(key=lambda x: x[0])
Bob.sort(key=lambda x: x[0])
Both.sort(key=lambda x: x[0])
none.sort(key=lambda x: x[0])
tresult = []
if 2 * k > m:
l = 2 * k - m
if len(Both) >= l:
tresult = Both[:l]
Both = Both[l:]
All = Alice + Both + Bob + none
m = 2 * (m - k)
k = k - l
else:
print(-1)
exit()
else:
tresult = []
tresult1 = []
if min(len(Alice), len(Bob)) == len(Alice):
if len(Alice) < k:
k1 = k - len(Alice)
if len(Both) < k1:
print(-1)
exit()
else:
tresult1 = Both[:k1]
Both = Both[k1:]
k = k - k1
elif len(Bob) < k:
k1 = k - len(Bob)
if len(Both) < k1:
print(-1)
exit()
else:
tresult1 = Both[:k1]
Both = Both[k1:]
k = k - k1
Alice1 = Alice[:k]
Bob1 = Bob[:k]
Alice = Alice[k:]
Bob = Bob[k:]
corr = []
calczz = m - 2 * k - len(tresult1)
if calczz > 0:
xtr = []
if len(Alice) > calczz:
xtr = Alice[:calczz]
else:
xtr = Alice
if len(Bob) > calczz:
xtr = xtr + Bob[:calczz]
else:
xtr = xtr + Bob
if len(none) > calczz:
xtr = xtr + none[:calczz]
else:
xtr = xtr + none
xtr = xtr[:calczz]
xtr.sort(key=lambda x: (x[1], x[2]), reverse=True)
zz = sum(row[1] == row[2] == 0 for row in xtr)
else:
zz = 0
if len(none) == zz:
nonechk = 9999999999
else:
nonechk = none[zz][0]
while (
len(Alice1) > 0
and len(Bob1) > 0
and len(Both) > 0
and len(none) > 0
and Alice1[-1][0] + Bob1[-1][0]
>= Both[0][0] + min(Alice1[-1][0], Bob1[-1][0], nonechk)
):
if min(Alice1[-1][0], Bob1[-1][0], nonechk) == nonechk:
zz += 1
if len(none) == zz:
nonechk = 9999999999
else:
nonechk = none[zz][0]
Alice.append(Alice1[-1])
Bob.append(Bob1[-1])
corr.append(Both[0])
Alice1.pop(-1)
Bob1.pop(-1)
Both.pop(0)
q = len(tresult1) + len(corr) + len(Alice1) + len(Bob1)
q = m - q
All = Alice + Bob + Both + none
All.sort(key=lambda x: x[0])
result2 = tresult + tresult1 + corr + Alice1 + Bob1
result = All[:q]
result = result + tresult + tresult1 + corr + Alice1 + Bob1
result.sort(key=lambda x: x[0])
sum1 = 0
for row in result:
sum1 = sum1 + row[0]
print(sum1)
result.sort(key=lambda x: x[3])
print(" ".join([str(row[3]) for row in result])) | IMPORT 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 ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR WHILE VAR ASSIGN VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR 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 LIST IF BIN_OP NUMBER VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST IF FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP NUMBER VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR LIST IF FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER VAR VAR ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR NUMBER WHILE FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER VAR IF FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR NUMBER 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 = map(int, input().split())
first = []
second = []
common = []
array = []
for s in range(n):
a, b, c = map(int, input().split())
array.append([a, b, c])
f, s, d = 0, 0, 0
for i in range(n):
a = array[i][0]
b = array[i][1]
c = array[i][2]
if b == 1 and c == 1:
common.append(a)
d += 1
elif b == 1 and c == 0:
first.append(a)
f += 1
elif b == 0 and c == 1:
second.append(a)
s += 1
mini = min(f, s)
first.sort()
second.sort()
for i in range(mini):
common.append(first[i] + second[i])
common.sort()
add = 0
if m <= d + mini:
for i in range(m):
add += common[i]
print(add)
else:
print(-1) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST 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 EXPR FUNC_CALL VAR LIST VAR VAR VAR ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER IF VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR 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 | from sys import stdin, stdout
n, k = stdin.readline().strip().split(" ")
n, k = int(n), int(k)
books = []
for i in range(n):
t, a, b = stdin.readline().strip().split(" ")
t, a, b = int(t), int(a), int(b)
books.append((t, a, b))
pairs_a = []
pairs_b = []
single = []
for i in books:
if i[1] == 1 and i[2] == 1:
single.append(i[0])
elif i[1] or i[2] == 1:
if i[1] == 1:
pairs_a.append(i[0])
else:
pairs_b.append(i[0])
pairs_b = sorted(pairs_b)
pairs_a = sorted(pairs_a)
pairs = []
for i in range(min(len(pairs_a), len(pairs_b))):
single.append(pairs_a[i] + pairs_b[i])
single = sorted(single)
if len(single) < k:
stdout.write(str(-1) + "\n")
else:
stdout.write(str(sum(single[:k])) + "\n") | ASSIGN VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR 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 VAR NUMBER NUMBER IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST 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 ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR NUMBER STRING EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR 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 | import sys
n, k = map(int, input().split())
a = []
b = []
g = []
for i in range(n):
t, x, y = map(int, input().split())
if x and y:
a.append(t)
elif x:
g.append(t)
elif y:
b.append(t)
a.sort()
b.sort()
g.sort()
t = 0
i = 0
j = 0
for o in range(k):
if i < len(a) and j < min(len(b), len(g)):
if a[i] < g[j] + b[j]:
t += a[i]
i += 1
else:
t += g[j] + b[j]
j += 1
elif i < len(a):
t += a[i]
i += 1
elif j < min(len(b), len(g)):
t += g[j] + b[j]
j += 1
else:
print(-1)
sys.exit(0)
print(t) | 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 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 FOR VAR FUNC_CALL 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 IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER IF VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL 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 | import sys
from itertools import accumulate
input = sys.stdin.readline
n, k = map(int, input().split())
BOOKS = [tuple(map(int, input().split())) for i in range(n)]
A = []
B = []
AB = []
for t, a, b in BOOKS:
if a == 1 and b == 1:
AB.append(t)
if a == 1 and b == 0:
A.append(t)
if a == 0 and b == 1:
B.append(t)
A.sort()
B.sort()
AB.sort()
SA = [0] + list(accumulate(A))
SB = [0] + list(accumulate(B))
SAB = [0] + list(accumulate(AB))
ANS = 1 << 31
for i in range(len(SAB)):
if 0 <= k - i < len(SA) and 0 <= k - i < len(SB):
ANS = min(ANS, SAB[i] + SA[k - i] + SB[k - i])
if ANS == 1 << 31:
print(-1)
else:
print(ANS) | 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 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 BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF NUMBER BIN_OP VAR VAR FUNC_CALL VAR VAR NUMBER 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 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 | import sys
def input():
return sys.stdin.readline().rstrip()
def input_split():
return [int(i) for i in input().split()]
n, k = input_split()
times = []
alice_likes = []
bob_likes = []
for _ in range(n):
t, a, b = input_split()
times.append(t)
alice_likes.append(a)
bob_likes.append(b)
if sum(alice_likes) < k or sum(bob_likes) < k:
ans = -1
else:
times_both = []
times_alice = []
times_bob = []
for book in range(n):
if alice_likes[book] == 1 and bob_likes[book] == 1:
times_both.append(times[book])
elif alice_likes[book] == 1:
times_alice.append(times[book])
elif bob_likes[book] == 1:
times_bob.append(times[book])
else:
pass
times_both.sort()
times_alice.sort()
times_bob.sort()
times_both = times_both + [100000] * (n - len(times_both))
times_alice = times_alice + [100000] * (n - len(times_alice))
times_bob = times_bob + [100000] * (n - len(times_bob))
ans = 0
p1, p2, p3 = 0, 0, 0
for i in range(k):
if times_both[p1] <= times_alice[p2] + times_bob[p3]:
ans += times_both[p1]
p1 += 1
else:
ans += times_alice[p2] + times_bob[p3]
p2 += 1
p3 += 1
print(ans) | IMPORT FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL 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 EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN 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 EXPR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR BIN_OP LIST NUMBER BIN_OP VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP LIST NUMBER BIN_OP VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP LIST NUMBER BIN_OP VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER FOR 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 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.buffer.readline
n, k = map(int, input().split())
a, b, c = [], [], []
for i in range(n):
t, x, y = map(int, input().split())
if x & y:
c.append(t)
elif x:
a.append(t)
elif y:
b.append(t)
a.sort(reverse=True)
b.sort(reverse=True)
c.sort(reverse=True)
ans = 0
for i in range(k):
if a and b and c:
if c[-1] < a[-1] + b[-1]:
ans += c.pop()
else:
ans += a.pop() + b.pop()
elif c:
ans += c.pop()
elif a and b:
ans += a.pop() + b.pop()
else:
exit(print(-1))
print(ans) | 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 BIN_OP 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 NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF 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 IF VAR VAR FUNC_CALL VAR IF VAR VAR VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR 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 | def solve():
n, k = list(map(int, input().split()))
a = []
b = []
both = []
alice = 0
bob = 0
coincidence = 0
for i in range(n):
t, a_, b_ = list(map(int, input().split()))
if b_ and a_:
both.append(t)
coincidence += 1
elif a_ == 1:
a.append(t)
alice += 1
elif b_ == 1:
b.append(t)
bob += 1
if alice + coincidence < k or bob + coincidence < k:
print(-1)
else:
a.sort()
b.sort()
for i in range(min(len(a), len(b))):
both.append(a[i] + b[i])
both.sort()
out = sum(both[:k])
print(out)
cases = 1
for test in range(cases):
solve() | FUNC_DEF ASSIGN VAR VAR FUNC_CALL 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 FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR 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 IF BIN_OP VAR VAR VAR BIN_OP 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 ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR 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 = map(int, input().split())
a, m, b = [], [], []
for i in range(n):
ti, ai, bi = map(int, input().split())
if ai == 1 and bi == 1:
m.append(ti)
elif ai == 1:
a.append(ti)
elif bi == 1:
b.append(ti)
a.sort()
m.sort()
b.sort()
pm = [0] * (len(m) + 1)
for i in range(1, len(m) + 1):
pm[i] = pm[i - 1] + m[i - 1]
pa = [0] * (len(a) + 1)
for i in range(1, len(a) + 1):
pa[i] = pa[i - 1] + a[i - 1]
pb = [0] * (len(b) + 1)
for i in range(1, len(b) + 1):
pb[i] = pb[i - 1] + b[i - 1]
ans = 10**12
for c in range(min(k + 1, len(m) + 1)):
if min(len(a), len(b)) >= k - c and ans > pm[c] + pa[k - c] + pb[k - c]:
ans = pm[c] + pa[k - c] + pb[k - c]
if ans == 10**12:
print(-1)
else:
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 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 FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP 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 BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP 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 BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP 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(" "))
x = [40000]
y = [40000]
z = [40000]
c = d = 0
for i in range(n):
t, a, b = map(int, input().split(" "))
if a == 1 and b == 1:
z.append(t)
c += 1
d += 1
elif a == 1:
x.append(t)
c += 1
elif b == 1:
y.append(t)
d += 1
if c < k or d < k:
print(-1)
else:
x.sort(reverse=True)
y.sort(reverse=True)
z.sort(reverse=True)
c = d = ans = 0
while c < k or d < k:
if c < k and d < k:
if x[len(x) - 1] + y[len(y) - 1] < z[len(z) - 1]:
ans += x[len(x) - 1] + y[len(y) - 1]
x.pop()
y.pop()
else:
ans += z[len(z) - 1]
z.pop()
c += 1
d += 1
elif c < k:
if x[len(x) - 1] < z[len(z) - 1]:
ans += x[len(x) - 1]
x.pop()
else:
ans += z[len(z) - 1]
z.pop()
c += 1
else:
if y[len(y) - 1] < z[len(z) - 1]:
ans += y[len(y) - 1]
y.pop()
else:
ans += z[len(z) - 1]
z.pop()
d += 1
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR 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 VAR NUMBER WHILE VAR VAR VAR VAR IF VAR VAR VAR VAR IF BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER VAR NUMBER IF VAR VAR IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL 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()]
books = [[int(i) for i in input().split()] for i in range(n)]
b01 = []
b10 = []
b11 = []
b00c, b01c, b10c, b11c = 0, 0, 0, 0
index = 0
for book in books:
if book[1] == 1:
if book[2] == 1:
b11.append(book[0])
else:
b10.append(book[0])
elif book[2] == 1:
b01.append(book[0])
b11.sort(), b10.sort(), b01.sort()
b0_index = 0
b1_index = 0
time = 0
if len(b11) > 0:
b1Out = False
else:
b1Out = True
if len(b01) > 0 and len(b10) > 0:
b0Out = False
else:
b0Out = True
while b0_index + b1_index < k:
if not any([b1Out, b0Out]):
if b11[b1_index] < b01[b0_index] + b10[b0_index]:
time += b11[b1_index]
b1_index += 1
if b1_index > len(b11) - 1:
b1Out = True
else:
time += b01[b0_index] + b10[b0_index]
b0_index += 1
if b0_index > len(b01) - 1 or b0_index > len(b10) - 1:
b0Out = True
elif b1Out and b0Out:
time = -1
break
elif b1Out:
time += b01[b0_index] + b10[b0_index]
b0_index += 1
if b0_index == len(b01) or b0_index == len(b10):
b0Out = True
else:
time += b11[b1_index]
b1_index += 1
if b1_index == len(b11):
b1Out = True
print(time) | ASSIGN VAR VAR FUNC_CALL VAR 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 LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR VAR VAR VAR NUMBER NUMBER NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER NUMBER IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR VAR IF FUNC_CALL VAR LIST VAR VAR IF VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER IF VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR FUNC_CALL 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 | n, k = list(map(int, input().split()))
ar1 = []
ar2 = []
ar3 = []
for i in range(n):
ar = list(map(int, input().split()))
if ar[1] == 1 and ar[2] == 0:
ar1.append(ar)
elif ar[1] == 0 and ar[2] == 1:
ar2.append(ar)
elif ar[1] == ar[2] == 1:
ar3.append(ar)
ar1.sort()
ar2.sort()
ar3.sort()
i, j, h = 0, 0, 0
num1, num2 = k, k
ans = 0
while (i < len(ar1) and j < len(ar2) or h < len(ar3)) and num1 > 0 and num2 > 0:
if h >= len(ar3):
ans += ar1[i][0] + ar2[j][0]
i += 1
j += 1
elif i >= len(ar1) or j >= len(ar2):
ans += ar3[h][0]
h += 1
elif ar1[i][0] + ar2[j][0] > ar3[h][0]:
ans += ar3[h][0]
h += 1
else:
ans += ar1[i][0] + ar2[j][0]
i += 1
j += 1
num1 -= 1
num2 -= 1
if num1 > 0 or num2 > 0:
print(-1)
else:
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 IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER 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 VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF BIN_OP VAR VAR NUMBER 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 VAR NUMBER VAR NUMBER 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 | n, k = map(int, input().split())
lForE = []
lForA = []
lForB = []
cnt = 0
forE = 0
forP = 0
ans = 0
for i in range(n):
temp = list(map(int, input().split()))
if temp[1] and temp[2]:
lForE.append(temp[0])
elif temp[1] and not temp[2]:
lForA.append(temp[0])
elif not temp[1] and temp[2]:
lForB.append(temp[0])
lForE.sort()
lForA.sort()
lForB.sort()
while cnt < k and (forP < len(lForB) and forP < len(lForA) or forE < len(lForE)):
cnt += 1
if forP < len(lForB) and forP < len(lForA) and forE < len(lForE):
if lForA[forP] + lForB[forP] < lForE[forE]:
ans += lForA[forP] + lForB[forP]
forP += 1
else:
ans += lForE[forE]
forE += 1
elif forE < len(lForE):
ans += lForE[forE]
forE += 1
elif forP < len(lForB) and forP < len(lForA):
ans += lForA[forP] + lForB[forP]
forP += 1
if cnt < k:
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 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 VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR WHILE VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER IF 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 VAR VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR 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 | import sys
input = lambda: sys.stdin.readline().rstrip()
inp = sys.stdin.buffer.readline
def I():
return list(map(int, inp().split()))
inf = 10**8
n, k = [int(i) for i in input().split()]
al = []
bo = []
bt = []
for i in range(n):
x = [int(i) for i in input().split()]
if x[1] == 1 and x[2] == 1:
bt.append(x[0])
elif x[1] == 0 and x[2] == 1:
bo.append(x[0])
elif x[1] == 1 and x[2] == 0:
al.append(x[0])
al.sort()
bo.sort()
bt.sort()
if len(bt) + len(al) < k or len(bt) + len(bo) < k:
print(-1)
exit(0)
else:
cnt = [[0, 0, 0] for i in range(max(len(al), len(bo), len(bt)))]
for i in range(len(cnt)):
if i < len(al):
cnt[i][0] = al[i]
else:
cnt[i][0] = inf
if i < len(bo):
cnt[i][1] = bo[i]
else:
cnt[i][1] = inf
if i < len(bt):
cnt[i][2] = bt[i]
else:
cnt[i][2] = inf
cnt.append([inf, inf, 2 * inf])
k1 = k
i = 0
j = 0
k = 0
val = 0
ans = 0
while val != k1:
if cnt[i][0] + cnt[j][1] >= cnt[k][2]:
ans += cnt[k][2]
val += 1
k += 1
else:
ans += cnt[i][1] + cnt[j][0]
val += 1
i += 1
j += 1
print(ans) | IMPORT ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP NUMBER NUMBER 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 FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL 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 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 NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER VAR VAR ASSIGN VAR VAR NUMBER VAR IF VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER VAR VAR ASSIGN VAR VAR NUMBER VAR IF VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER VAR VAR ASSIGN VAR VAR NUMBER VAR EXPR FUNC_CALL VAR LIST VAR VAR BIN_OP NUMBER VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF BIN_OP VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR NUMBER 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 | n, k = map(int, input().split())
a = []
b = []
c = []
for i in range(n):
t, x, y = map(int, input().split())
if x:
if y:
a.append(t)
else:
c.append(t)
elif y:
b.append(t)
if len(a) + len(b) < k or len(a) + len(c) < k:
print(-1)
else:
mn = 10**10
a = sorted(a)
b = sorted(b)
c = sorted(c)
apre = [0]
cpre = [0]
bpre = [0]
for num in a:
apre.append(apre[-1] + num)
for num in b:
bpre.append(bpre[-1] + num)
for num in c:
cpre.append(cpre[-1] + num)
for tk in range(min(k + 1, len(a) + 1)):
chane = k - tk
if chane <= len(b) and chane <= len(c):
sm = apre[tk] + bpre[chane] + cpre[chane]
mn = min(sm, mn)
print(mn) | 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 IF VAR EXPR FUNC_CALL VAR 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 ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER FOR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FOR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FOR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL 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 binary(a, start, end, e):
mid = (start + end) // 2
if a[mid] == e:
return mid
elif a[mid] > e:
end = mid - 1
else:
start = mid + 1
if start <= end:
return binary(a, start, end, e)
else:
return start
n, k = map(int, input().split())
p = []
al = []
bo = []
for j in range(n):
a, b, c = map(int, input().split())
if b == 1 and c == 1:
p.append(a)
elif b == 1 and c == 0:
al.append(a)
elif b == 0 and c == 1:
bo.append(a)
al.sort()
bo.sort()
if p != []:
while al != [] and bo != []:
v = al[0] + bo[0]
p.append(v)
del al[0]
del bo[0]
else:
while al != [] and bo != []:
v = al[0] + bo[0]
p.append(v)
del al[0]
del bo[0]
p.sort()
if len(p) < k:
print(-1)
else:
print(sum(p[:k])) | FUNC_DEF ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR RETURN VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR VAR RETURN 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 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 IF VAR LIST WHILE VAR LIST VAR LIST ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER WHILE VAR LIST VAR LIST ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER 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 | def min_time(tot_books, books_like, read_time, a_time, b_time):
time = []
temp_a = []
temp_b = []
if min(sum(a_time), sum(b_time)) >= books_like:
for x in range(tot_books):
if a_time[x] == b_time[x] == 1:
time.append(read_time[x])
elif a_time[x] == 0 and b_time[x] == 1:
temp_b.append(read_time[x])
elif a_time[x] == 1 and b_time[x] == 0:
temp_a.append(read_time[x])
temp_a.sort(), temp_b.sort()
for y in range(min(len(temp_a), len(temp_b))):
time.append(temp_a[y] + temp_b[y])
time.sort()
time = time[:books_like]
return sum(time)
else:
return -1
n, k = map(int, input().split())
t = []
a = []
b = []
for i in range(n):
x, y, z = map(int, input().split())
t.append(x), a.append(y), b.append(z)
print(min_time(n, k, t, a, b)) | FUNC_DEF ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST IF FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR 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 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 ASSIGN VAR VAR VAR RETURN FUNC_CALL VAR VAR RETURN NUMBER 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 EXPR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR 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, r = map(int, input().split())
l = []
for _ in range(n):
l.append(list(map(int, input().split())))
l1, l2, l3 = [], [], []
for i in l:
if i[1] == 1 and i[2] == 1:
l1.append(i)
elif i[1] == 1:
l2.append(i)
elif i[2] == 1:
l3.append(i)
l1.sort(key=lambda x: x[0])
l2.sort(key=lambda x: x[0])
l3.sort(key=lambda x: x[0])
n1, n2, n3 = len(l1), len(l2), len(l3)
i, j, k = 0, 0, 0
a, b, s = 0, 0, 0
ans = 0
while a < r or b < r:
if a < r and b < r:
if i < n1 and (j < n2 and k < n3):
if s + l1[i][0] < s + l2[j][0] + l3[k][0]:
s += l1[i][0]
i += 1
a += 1
b += 1
else:
s += l2[j][0] + l3[k][0]
j += 1
k += 1
a += 1
b += 1
elif i >= n1 and (j < n2 and k < n3):
s += l2[j][0] + l3[k][0]
j += 1
k += 1
a += 1
b += 1
elif i < n1 and (j >= n2 or k >= n3):
s += l1[i][0]
i += 1
a += 1
b += 1
else:
ans = -1
break
elif a < r:
if i < n1 and j < n2:
if s + l1[i][0] < s + l2[j][0]:
s += l1[i][0]
i += 1
a += 1
else:
s += l2[j][0]
j += 1
a += 1
elif i > n1 and j < n2:
s += l2[j][0]
j += 1
a += 1
elif j > n2 and i < n1:
s + l1[i][0]
i += 1
a += 1
else:
ans = -1
break
elif i < n1 and k < n3:
if s + l1[i][0] < s + l[k][0]:
s += l1[i][0]
i += 1
b += 1
else:
s += l3[k][0]
k += 1
b += 1
elif i > n1 and k < n3:
s += l3[k][0]
k += 1
b += 1
elif k > n3 and i < n1:
s + l1[i][0]
i += 1
b += 1
else:
ans = -1
break
print(s if a == r and b == r else -1) | 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 VAR VAR LIST LIST LIST FOR VAR VAR IF VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER NUMBER 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 VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR IF VAR VAR VAR VAR IF VAR VAR VAR VAR VAR VAR IF BIN_OP VAR VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR NUMBER VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR IF VAR VAR VAR VAR IF BIN_OP VAR VAR VAR NUMBER BIN_OP VAR VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR EXPR BIN_OP VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR VAR IF BIN_OP VAR VAR VAR NUMBER BIN_OP VAR VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR EXPR BIN_OP VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR 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 = list(map(int, input().split()))
al = []
bl = []
all = []
for bb in range(n):
t, a, b = list(map(int, input().split()))
if a == 1:
if b == 1:
all.append(t)
else:
al.append(t)
elif b == 1:
bl.append(t)
ml = min(len(al), len(bl))
if len(all) + ml < k:
print(-1)
else:
al = sorted(al)
bl = sorted(bl)
for j in range(ml):
all.append(al[j] + bl[j])
all = sorted(all)
print(sum(all[:k])) | 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 IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR 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 | N, K = [int(i) for i in input().split()]
T, A, B = [], [], []
Tall = []
Talice = []
Tbob = []
for i in range(N):
t, a, b = [int(j) for j in input().split()]
T.append(t)
A.append(a)
B.append(b)
if a and b:
Tall.append(t)
elif a:
Talice.append(t)
elif b:
Tbob.append(t)
Tall = sorted(Tall)
Talice = sorted(Talice)
Tbob = sorted(Tbob)
iall = 0
isep = 0
ans = 0
ml = min(len(Talice), len(Tbob))
for i in range(K):
if iall >= len(Tall) and isep >= ml:
ans = -1
break
if iall < len(Tall) and (isep >= ml or Tall[iall] <= Talice[isep] + Tbob[isep]):
ans += Tall[iall]
iall += 1
else:
ans += Talice[isep] + Tbob[isep]
isep += 1
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR LIST LIST LIST 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 EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR IF VAR 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 NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR NUMBER 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 = map(int, input().split())
cnta = []
cntb = []
cntab = []
for i in range(n):
t, a, b = map(int, input().split())
if a:
if b:
cntab.append(t)
else:
cnta.append(t)
elif b:
cntb.append(t)
T = 0
cnta = sorted(cnta)
cntb = sorted(cntb)
cntab = sorted(cntab)
l, m = 0, 0
r = min(len(cnta), len(cntb))
it = 0
while l < r and it < k:
if m >= len(cntab) or cnta[l] + cntb[l] < cntab[m]:
T += cnta[l] + cntb[l]
l += 1
else:
T += cntab[m]
m += 1
it += 1
if it + len(cntab) - m < k:
print(-1)
else:
print(T + sum(cntab[m : m + k - it])) | 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 IF VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN 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 VAR VAR VAR NUMBER VAR NUMBER IF BIN_OP BIN_OP VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL 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 main():
n, k = [int(x) for x in input().split()]
alice, bob = 0, 0
both, j_a, j_b = [], [], []
for i in range(n):
b = [int(x) for x in input().split()]
if b[1] + b[2] == 2:
both.append(b)
elif b[1] == 1:
j_a.append(b)
elif b[2] == 1:
j_b.append(b)
j_a.sort()
j_b.sort()
for i in range(min(len(j_a), len(j_b))):
b = [j_a[i][0] + j_b[i][0], 1, 1]
both.append(b)
if len(both) < k:
return -1
both.sort()
return sum([b[0] for b in both[:k]])
print(main()) | FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP VAR NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST BIN_OP VAR VAR NUMBER VAR VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR RETURN NUMBER EXPR FUNC_CALL VAR RETURN FUNC_CALL VAR VAR NUMBER VAR 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 | def find(books, k):
both = []
A = []
B = []
for t, a, b in books:
if a == 1 and b == 1:
both += [t]
elif a == 1:
A += [t]
elif b == 1:
B += [t]
both = sorted(both)
A = sorted(A)
B = sorted(B)
ans = 10**10
num = 0
temp = 0
pre_both = []
for i in range(len(both)):
temp += both[i]
pre_both += [temp]
temp = 0
pre_A = []
for i in range(len(A)):
temp += A[i]
pre_A += [temp]
temp = 0
pre_B = []
for i in range(len(B)):
temp += B[i]
pre_B += [temp]
for num in range(min(k, len(pre_both)) + 1):
if num > len(pre_both):
break
if num > 0:
need = pre_both[num - 1]
if num == k:
ans = min(ans, need)
break
remain = k - num
if remain > len(A):
continue
if remain > len(B):
continue
if num > 0:
need = pre_both[num - 1] + pre_A[remain - 1] + pre_B[remain - 1]
else:
need = pre_A[remain - 1] + pre_B[remain - 1]
ans = min(ans, need)
if ans == 10**10:
return -1
return ans
n, k = list(map(int, input().strip().split()))
books = []
for _ in range(n):
t, a, b = list(map(int, input().strip().split()))
books += [(t, a, b)]
print(find(books, k)) | FUNC_DEF ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER VAR LIST VAR IF VAR NUMBER VAR LIST VAR IF VAR NUMBER VAR LIST VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR VAR LIST VAR ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR VAR LIST VAR ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR VAR LIST VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR BIN_OP NUMBER NUMBER RETURN NUMBER RETURN VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR VAR LIST VAR VAR 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
INF = 10**18
n, m, k = map(int, input().split())
B = [tuple(map(int, input().split())) for _ in range(n)]
GB = []
AB = []
BB = []
RB = []
for i, (t, a, b) in enumerate(B):
if a and b:
GB.append((t, i))
elif a:
AB.append((t, i))
elif b:
BB.append((t, i))
else:
RB.append((t, i))
GB.sort()
AB.sort()
BB.sort()
CB = sorted((t1 + t2, i1, i2) for (t1, i1), (t2, i2) in zip(AB, BB))
N = 1
while N <= 10**4:
N *= 2
T = [(0, 0)] * (2 * N)
def comb(a, b):
return a[0] + b[0], a[1] + b[1]
def add_book(t, inc=1):
i = t + N
T[i] = comb(T[i], (t * inc, inc))
while i > 1:
i //= 2
T[i] = comb(T[i * 2], T[i * 2 + 1])
def query(x):
assert x >= 0
s = 0
i = 1
while x:
ts, tc = T[i]
if tc < x:
return INF
elif tc == x:
s += ts
break
if i >= N:
s += ts // tc * x
break
i *= 2
if T[i][1] < x:
s += T[i][0]
x -= T[i][1]
i += 1
return s
for t, _ in RB:
add_book(t)
for t, _ in AB:
add_book(t)
for t, _ in BB:
add_book(t)
gb_i = 0
gb_t = 0
while gb_i < min(len(GB), m):
gb_t += GB[gb_i][0]
gb_i += 1
for t, _ in GB[gb_i:]:
add_book(t)
cb_i = 0
cb_t = 0
while gb_i + cb_i < k and gb_i + 2 * (cb_i + 1) <= m and cb_i < len(CB):
cb_t += CB[cb_i][0]
add_book(AB[cb_i][0], -1)
add_book(BB[cb_i][0], -1)
cb_i += 1
if gb_i + cb_i < k:
print(-1)
sys.exit()
best = INF, -1, -1
while True:
best = min(best, (gb_t + cb_t + query(m - 2 * cb_i - gb_i), gb_i, cb_i))
if not gb_i:
break
gb_i -= 1
gb_t -= GB[gb_i][0]
add_book(GB[gb_i][0])
if gb_i + cb_i < k:
if cb_i == len(CB):
break
cb_t += CB[cb_i][0]
add_book(AB[cb_i][0], -1)
add_book(BB[cb_i][0], -1)
cb_i += 1
if gb_i + 2 * cb_i > m:
break
bs, bi, bj = best
assert bs != INF
ans = []
comps = 0
for t, i in GB[:bi]:
ans.append(i + 1)
comps += t
for t, i1, i2 in CB[:bj]:
ans.append(i1 + 1)
ans.append(i2 + 1)
comps += t
if bi + 2 * bj < m:
rem = GB[bi:] + AB[bj:] + BB[bj:] + RB
rem.sort()
for t, i in rem[: m - bi - 2 * bj]:
ans.append(i + 1)
comps += t
assert comps == bs
print(bs)
print(*ans) | IMPORT ASSIGN VAR VAR ASSIGN VAR BIN_OP NUMBER 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 VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR VAR VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR EXPR FUNC_CALL VAR VAR VAR IF VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR BIN_OP NUMBER NUMBER VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER BIN_OP NUMBER VAR FUNC_DEF RETURN BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER FUNC_DEF NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR WHILE VAR NUMBER VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER FUNC_DEF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR VAR VAR VAR IF VAR VAR RETURN VAR IF VAR VAR VAR VAR IF VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR NUMBER IF VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER RETURN VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER FOR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR VAR BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER VAR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR NUMBER IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER WHILE NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR BIN_OP NUMBER VAR VAR VAR VAR IF VAR VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER IF BIN_OP VAR VAR VAR IF VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR NUMBER IF BIN_OP VAR BIN_OP NUMBER VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR FOR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR IF BIN_OP VAR BIN_OP NUMBER VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR FOR VAR VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR VAR 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 | n, k = map(int, input().split())
t11 = []
t01 = []
t10 = []
for j in range(n):
ti, ai, bi = map(int, input().split())
if ai == 0 and bi == 0:
continue
elif ai == 1 and bi == 1:
t11.append(ti)
elif ai == 0 and bi == 1:
t01.append(ti)
else:
t10.append(ti)
lt11 = len(t11)
lt10 = len(t10)
lt01 = len(t01)
if lt11 + min(lt10, lt01) >= k:
t01 = sorted(t01)
t10 = sorted(t10)
t1 = t0 = tot = 0
for i in range(min(lt10, lt01)):
t11.append(t10[i] + t01[i])
print(sum(sorted(t11)[:k]))
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 IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR 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 FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR 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, *f = map(int, open(0).read().split())
books = [f[i * 3 : i * 3 + 3] for i in range(n)]
a = []
b = []
ab = []
ans = float("inf")
for x in books:
if x[1] == 1:
if x[2] == 1:
ab.append(x[0])
else:
a.append(x[0])
elif x[2] == 1:
b.append(x[0])
m = min(len(a), len(b))
if m + len(ab) >= k:
a.sort()
b.sort()
ab.sort()
csa = [0]
for i in range(len(a)):
csa.append(csa[i] + a[i])
csb = [0]
for i in range(len(b)):
csb.append(csb[i] + b[i])
csab = [0]
for i in range(len(ab)):
csab.append(csab[i] + ab[i])
for i in range(max(0, k - m), min(k, len(ab)) + 1):
ans = min(ans, csab[i] + csa[k - i] + csb[k - i])
print(ans)
else:
print(-1) | ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR STRING FOR VAR VAR IF VAR NUMBER NUMBER IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER 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 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())
l = []
for i in range(n):
l.append(list(map(int, input().split())))
x, y, z = [], [], []
for i in range(n):
if l[i][1] == 1 and l[i][2] == 1:
x.append(l[i][0])
elif l[i][1] == 1 and l[i][2] == 0:
y.append(l[i][0])
elif l[i][1] == 0 and l[i][2] == 1:
z.append(l[i][0])
y.sort()
z.sort()
for i in range(min(len(y), len(z))):
x.append(y[i] + z[i])
x.sort()
if len(x) < k:
print(-1)
else:
print(sum(x[:k])) | 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 VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR 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 IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER 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 | n, k = [int(x) for x in input().strip().split()]
vec = []
for _ in range(n):
vec.append([int(x) for x in input().split()])
vec.sort()
f1 = f2 = f3 = 0
minsum = 0
count = 0
min_12 = min_3 = 0
flag12 = flag3 = True
while (f3 <= n - 1 or f1 <= n - 1 and f2 <= n - 1) and count < k:
while f3 <= n - 1:
if vec[f3][1] == 1 and vec[f3][2] == 1:
break
f3 += 1
if f3 <= n - 1:
min_3 = vec[f3][0]
else:
flag3 = False
while f2 <= n - 1:
if vec[f2][1] == 1 and vec[f2][2] == 0:
break
f2 += 1
while f1 <= n - 1:
if vec[f1][1] == 0 and vec[f1][2] == 1:
break
f1 += 1
if f1 <= n - 1 and f2 <= n - 1:
min_12 = vec[f1][0] + vec[f2][0]
else:
flag12 = False
if flag12 and flag3:
count += 1
if min_3 <= min_12:
f3 += 1
minsum += min_3
else:
f1 += 1
f2 += 1
minsum += min_12
elif flag12:
count += 1
f1 += 1
f2 += 1
minsum += min_12
elif flag3:
count += 1
f3 += 1
minsum += min_3
if count == k:
print(minsum)
else:
print(-1) | ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL 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 EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER WHILE VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR WHILE VAR BIN_OP VAR NUMBER IF VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR NUMBER IF VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER VAR NUMBER WHILE VAR BIN_OP VAR NUMBER IF VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER VAR VAR IF VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR VAR IF VAR VAR NUMBER VAR NUMBER VAR VAR 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 | from sys import stdin
input = stdin.readline
INF = 10**9 + 7
MAX = 10**7 + 7
MOD = 10**9 + 7
n, k = [int(x) for x in input().strip().split()]
c, a, b = [], [], []
for ni in range(n):
ti, ai, bi = [int(x) for x in input().strip().split()]
if ai == 1 and bi == 1:
c.append(ti)
elif ai == 1:
a.append(ti)
elif bi == 1:
b.append(ti)
c.sort(reverse=True)
a.sort(reverse=True)
b.sort(reverse=True)
alen = len(a)
blen = len(b)
clen = len(c)
m = max(0, k - min(alen, blen))
ans = 0
if m > clen:
print("-1")
else:
for mi in range(m):
ans += c.pop()
ka = k - m
kb = k - m
while ka or kb:
ca = c[-1] if c else float("inf")
da = 0
ap, bp = 0, 0
if ka:
da += a[-1] if a else float("inf")
ap = 1
if kb:
da += b[-1] if b else float("inf")
bp = 1
if da < ca:
if ap:
ka -= 1
ans += a.pop() if a else float("inf")
if bp:
kb -= 1
ans += b.pop() if b else float("inf")
else:
ans += c.pop() if c else float("inf")
if ap:
ka -= 1
if bp:
kb -= 1
print(ans if ans != float("inf") else "-1") | ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL 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 VAR FUNC_CALL 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 NUMBER EXPR FUNC_CALL VAR NUMBER 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 FUNC_CALL VAR NUMBER BIN_OP VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR WHILE VAR VAR ASSIGN VAR VAR VAR NUMBER FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER IF VAR VAR VAR VAR NUMBER FUNC_CALL VAR STRING ASSIGN VAR NUMBER IF VAR VAR VAR VAR NUMBER FUNC_CALL VAR STRING ASSIGN VAR NUMBER IF VAR VAR IF VAR VAR NUMBER VAR VAR FUNC_CALL VAR FUNC_CALL VAR STRING IF VAR VAR NUMBER VAR VAR FUNC_CALL VAR FUNC_CALL VAR STRING VAR VAR FUNC_CALL VAR FUNC_CALL VAR STRING IF VAR VAR NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR FUNC_CALL VAR STRING 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 | import sys
def fun(x):
return x[0]
n, k = list(map(int, sys.stdin.readline().strip().split()))
xx = []
a = []
b = []
c = []
for i in range(n):
x = list(map(int, sys.stdin.readline().strip().split()))
if x[1] == x[2] == 1:
a.append(x[0])
elif x[1] == 1:
b.append(x[0])
elif x[2] == 1:
c.append(x[0])
b.sort()
c.sort()
for i in range(min(len(b), len(c))):
a.append(b[i] + c[i])
a.sort()
if len(a) < k:
print(-1)
else:
print(sum(a[:k])) | IMPORT FUNC_DEF RETURN VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST 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 FUNC_CALL VAR IF VAR 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 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 IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER 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 | n, k = list(map(int, input().split()))
both = []
alice = []
bob = []
for i in range(n):
t, a, b = list(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()
x = 0
y = 0
if len(both) + len(alice) < k or len(both) + len(bob) < k:
print(-1)
else:
both.append(10**9)
alice.append(10**9)
bob.append(10**9)
i = 0
j = 0
time = 0
alpha = 0
while x < k and y < k:
if i < len(both) and j < len(alice) and alpha < len(bob):
if both[i] <= alice[j] + bob[alpha]:
time += both[i]
i += 1
x += 1
y += 1
else:
time += alice[j] + bob[alpha]
j += 1
alpha += 1
x += 1
y += 1
elif len(both) != i:
if len(alice) == 0:
if len(bob) == 0:
time += both[i]
i += 1
x += 1
y += 1
print(time) | 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 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 EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR 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 IF VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER 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 | import itertools
n, k = map(int, input().split())
both = [0]
alice = [0]
bob = [0]
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)
exit()
both.sort()
alice.sort()
bob.sort()
cs_both = list(itertools.accumulate(both))
cs_alice = list(itertools.accumulate(alice))
cs_bob = list(itertools.accumulate(bob))
ans = float("inf")
for i in range(k + 1):
try:
tmp = cs_both[i]
except IndexError:
break
if i == k:
ans = min(ans, tmp)
try:
tmp += cs_alice[k - i]
except IndexError:
continue
try:
tmp += cs_bob[k - i]
except IndexError:
continue
ans = min(ans, tmp)
if ans == float("inf"):
print(-1)
else:
print(ans) | IMPORT ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR LIST 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 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 EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR FUNC_CALL VAR STRING 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(" "))
book = [[] for i in range(4)]
s = [[] for i in range(4)]
for i in range(n):
t, a, b = map(int, input().split(" "))
book[2 * a + b].append(t)
for i in range(1, 4):
book[i].sort()
s[i].append(0)
for j in book[i]:
s[i].append(s[i][len(s[i]) - 1] + j)
ans = int(2000000000.0 + 1)
for i in range(min(k + 1, len(s[3]))):
if k - i < len(s[1]) and k - i < len(s[2]):
ans = min(ans, s[3][i] + s[1][k - i] + s[2][k - i])
print(-1 if ans == int(2000000000.0 + 1) else ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NUMBER FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP NUMBER 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 EXPR FUNC_CALL VAR VAR FUNC_CALL 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 = [int(x) for x in input().split()]
alice, bob, both = [], [], []
for i in range(n):
t, a, b = [int(x) for x in input().split()]
if a == 0 and b == 0:
continue
if a & 1 and b & 1:
both.append(t)
continue
if a & 1:
alice.append(t)
else:
bob.append(t)
alice.sort()
bob.sort()
for i in range(min(len(alice), len(bob))):
both.append(alice[i] + bob[i])
if len(both) < k:
print(-1)
else:
print(sum(sorted(both)[:k])) | ASSIGN VAR VAR FUNC_CALL VAR 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 VAR FUNC_CALL FUNC_CALL VAR IF VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR 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 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 | n, k = map(int, input().split())
oo = list()
oa = list()
ob = list()
for i in range(n):
t, a, b = map(int, input().split())
if a == 1 and b == 1:
oo.append(t)
elif a == 0 and b == 1:
ob.append(t)
elif a == 1 and b == 0:
oa.append(t)
oo = sorted(oo)
oa = sorted(oa)
ob = sorted(ob)
oo_p = 0
oa_p = 0
ob_p = 0
ca = 0
cb = 0
ans = 0
MAX = 23942034809238409823048
if max(0, max(k - len(oa), k - len(ob))) > len(oo):
print("-1")
exit(0)
def get_first_elem_from_list(l, pos):
if pos < len(l):
return l[pos]
else:
return MAX
def remove_first_elem_from_list(l, pos):
if len(l) > pos:
pos += 1
return pos
while ca < k or cb < k:
oo_f = get_first_elem_from_list(oo, oo_p)
oa_f = get_first_elem_from_list(oa, oa_p)
ob_f = get_first_elem_from_list(ob, ob_p)
if ca < k and cb < k:
if oo_f <= oa_f + ob_f:
ca += 1
cb += 1
ans += oo_f
oo_p = remove_first_elem_from_list(oo, oo_p)
elif oa_f + ob_f < oo_f:
ca += 1
cb += 1
ans += oa_f + ob_f
oa_p = remove_first_elem_from_list(oa, oa_p)
ob_p = remove_first_elem_from_list(ob, ob_p)
elif ca < k:
if oo_f <= oa_f:
ca += 1
ans += oo_f
oo_p = remove_first_elem_from_list(oo, oo_p)
elif oa_f < oo_f:
ca += 1
ans += oa_f
oa_p = remove_first_elem_from_list(oa, oa_p)
elif oo_f <= ob_f:
cb += 1
ans += oo_f
oo_p = remove_first_elem_from_list(oo, oo_p)
elif ob_f < oo_f:
cb += 1
ans += ob_f
ob_p = remove_first_elem_from_list(ob, ob_p)
print(ans) | ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR 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 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 ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR NUMBER FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR BIN_OP VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER FUNC_DEF IF VAR FUNC_CALL VAR VAR RETURN VAR VAR RETURN VAR FUNC_DEF IF FUNC_CALL VAR VAR VAR VAR NUMBER RETURN VAR WHILE VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR IF VAR BIN_OP VAR VAR VAR NUMBER VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR VAR VAR VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR IF VAR VAR VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL 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 cumsum(x):
n = len(x)
ans = [0] * (n + 1)
for i in range(n):
ans[i + 1] = ans[i] + x[i]
return ans
def solve(k, t, a, b):
n = len(t)
b10, b01, b11 = [], [], []
for i in range(n):
if a[i] == 1 and b[i] == 1:
b11.append(t[i])
elif a[i] == 1:
b10.append(t[i])
elif b[i] == 1:
b01.append(t[i])
b10 = sorted(b10)
b01 = sorted(b01)
b11 = sorted(b11)
b1 = [(b10[i] + b01[i]) for i in range(min(len(b10), len(b01)))]
cs_b1 = cumsum(b1)
cs_b11 = cumsum(b11)
if len(b1) + len(b11) < k:
return -1
ans = 1e100
for i in range(0, k + 1):
if i <= len(b1) and k - i <= len(b11):
ans = min(ans, cs_b1[i] + cs_b11[k - i])
return ans
n, k = map(int, input().split())
a, b, t = [0] * n, [0] * n, [0] * n
for i in range(n):
t[i], a[i], b[i] = map(int, input().split())
print(solve(k, t, a, b)) | FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR 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 VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR 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 RETURN NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR RETURN VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR BIN_OP LIST NUMBER VAR BIN_OP LIST NUMBER VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR 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 | l = input().split()
n = int(l[0])
k = int(l[1])
l = []
arrboth = []
arralice = []
arrbob = []
for i in range(n):
lo = input().split()
x = int(lo[0])
u = int(lo[1])
v = int(lo[2])
if u == 1 and v == 1:
arrboth.append(x)
elif u == 1:
arralice.append(x)
elif v == 1:
arrbob.append(x)
arrbob.sort()
arralice.sort()
arrboth.sort()
pref1 = [(0) for i in range(len(arrboth) + 1)]
pref2 = [(0) for i in range(len(arralice) + 1)]
pref3 = [(0) for i in range(len(arrbob) + 1)]
sumi = 0
for i in range(1, len(arrboth) + 1):
sumi += arrboth[i - 1]
pref1[i] = sumi
sumi = 0
for i in range(1, len(arrbob) + 1):
sumi += arrbob[i - 1]
pref3[i] = sumi
sumi = 0
for i in range(1, len(arralice) + 1):
sumi += arralice[i - 1]
pref2[i] = sumi
z = []
for i in range(min(len(arrboth), k) + 1):
if k - i > len(arrbob) or k - i > len(arralice):
continue
z.append(pref1[i] + pref2[k - i] + pref3[k - i])
if z == []:
print(-1)
else:
print(min(z)) | ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR 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 NUMBER VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR LIST 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 EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR IF VAR LIST 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 | import sys
input = sys.stdin.readline
def solve(n, k, t, a, b):
alice_only = []
bob_only = []
both = []
for i in range(n):
if a[i] and b[i]:
both.append(t[i])
elif a[i] and not b[i]:
alice_only.append(t[i])
elif b[i] and not a[i]:
bob_only.append(t[i])
if len(bob_only) + len(both) < k:
return -1
elif len(alice_only) + len(both) < k:
return -1
bob_only.sort()
alice_only.sort()
both.sort()
both_pre = [0]
bob_pre = [0]
alice_pre = [0]
for tb in bob_only:
bob_pre.append(bob_pre[-1] + tb)
for ta in alice_only:
alice_pre.append(alice_pre[-1] + ta)
for tbth in both:
both_pre.append(both_pre[-1] + tbth)
best = float("inf")
lb1 = k - len(alice_only)
lb2 = k - len(bob_only)
for cur_k in range(max(lb1, lb2, 0), min(len(both_pre), k + 1)):
time_both = both_pre[cur_k]
time_alice = alice_pre[k - cur_k]
time_bob = bob_pre[k - cur_k]
best = min(best, time_both + time_alice + time_bob)
if len(alice_pre) > k and len(bob_pre) > k:
best = min(best, alice_pre[k] + bob_pre[k])
return best
n, k = map(int, input().split())
t, a, b = [], [], []
for _ in range(n):
ti, ai, bi = map(int, input().split())
t.append(ti)
a.append(ai)
b.append(bi)
print(solve(n, k, t, a, b)) | IMPORT ASSIGN VAR VAR FUNC_DEF 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 VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR RETURN NUMBER IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR RETURN NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER FOR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FOR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FOR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP 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 EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR 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())
times = [[], [], [], []]
for i in range(n):
t, a, b = map(int, input().split())
times[a * 2 + b].append(t)
for i in range(1, 4):
times[i].sort()
sums = []
for i in range(len(times)):
sums.append([])
for j in range(len(times[i])):
if j == 0:
sums[i].append(times[i][0])
else:
sums[i].append(times[i][j] + sums[i][j - 1])
sums[1].insert(0, 0)
sums[2].insert(0, 0)
sums[3].insert(0, 0)
ans = 10**10
for i in range(min(k + 1, len(sums[-1]))):
if k - i < len(sums[1]) and k - i < len(sums[2]):
ans = min(ans, sums[3][i] + sums[1][k - i] + sums[2][k - i])
if ans == 10**10:
ans = -1
print(ans) | 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 FOR VAR FUNC_CALL VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER 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 BIN_OP NUMBER 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 | def fn(qa, qb, qc, k):
qa.sort()
qb.sort()
qc.sort()
num = 0
ia = ib = ic = 0
na, nb, nc = len(qa), len(qb), len(qc)
for i in range(k):
if nc == 0 or ic >= nc:
num += qa[ia] + qb[ib]
ia += 1
ib += 1
elif na == 0 or nb == 0 or ia >= na or ib >= nb:
num += qc[ic]
ic += 1
elif qa[ia] + qb[ib] < qc[ic]:
num += qa[ia] + qb[ib]
ia += 1
ib += 1
else:
num += qc[ic]
ic += 1
print(num)
n, k = list(map(int, input().split(" ")))
qa, qb, qc = [], [], []
ca = cb = 0
for i in range(n):
t, a, b = list(map(int, input().split(" ")))
if a == 1 and b == 1:
ca += 1
cb += 1
qc.append(t)
elif a == 1:
ca += 1
qa.append(t)
elif b == 1:
cb += 1
qb.append(t)
if ca < k or cb < k:
print(-1)
else:
fn(qa, qb, qc, k) | FUNC_DEF EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR VAR VAR 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 EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR VAR LIST LIST LIST ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING IF VAR NUMBER VAR NUMBER 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 VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR 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())
a = []
b = []
c = []
for i in range(n):
q, w, e = map(int, input().split())
if w and e:
a.append(q)
elif w:
b.append(q)
elif e:
c.append(q)
if len(a) + len(b) < k or len(a) + len(c) < k:
print(-1)
else:
a.sort()
b.sort()
c.sort()
w = 0
t = 0
o = 0
p = 0
while w < k:
if len(b) > o and len(c) > o and (p == len(a) or c[o] + b[o] < a[p]):
t = t + c[o] + b[o]
o = o + 1
else:
t = t + a[p]
p = p + 1
w = w + 1
print(t) | 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 NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP 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 | n, k = map(int, input().split())
A, B, both = [0], [0], [0]
for i in range(n):
t, a, b = map(int, input().split())
if a == 1 and b == 1:
both.append(t)
if a == 1 and b == 0:
A.append(t)
if a == 0 and b == 1:
B.append(t)
A.sort()
B.sort()
both.sort()
if len(A) - 1 + len(both) - 1 < k or len(B) - 1 + len(both) - 1 < k:
print(-1)
exit(0)
for i in range(1, len(A)):
A[i] += A[i - 1]
for i in range(1, len(B)):
B[i] += B[i - 1]
cur = 0
ans = 10**12 + 10
for ch in range(len(both)):
cur += both[ch]
if ch > k:
break
if k - ch < len(A) and k - ch < len(B):
ans = min(ans, A[k - ch] + B[k - ch] + cur)
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