time limit per test
2 seconds
memory limit per test
256 megabytes
Victor wants to become "Mr. Perfectly Fine". For that, he needs to acquire a certain set of skills. More precisely, he has 22 skills he needs to acquire.
Victor has nn books. Reading book ii takes him mimi minutes and will give him some (possibly none) of the required two skills, represented by a binary string of length 22.
What is the minimum amount of time required so that Victor acquires all of the two skills?
Input
The input consists of multiple test cases. The first line contains an integer tt (1≤t≤10001≤t≤1000) --- the number of test cases. The description of the test cases follows.
The first line of each test case contains an integer nn (1≤n≤2⋅1051≤n≤2⋅105) --- the number of books available.
Then nn lines follow. Line ii contains a positive integer mimi (1≤mi≤2⋅1051≤mi≤2⋅105) and a binary string of length 22, where si1=1si1=1 if reading book ii acquires Victor skill 11, and si1=0si1=0 otherwise, and si2=1si2=1 if reading book ii acquires Victor skill 22, and si2=0si2=0 otherwise.
It is guaranteed that the sum of nn over all test cases doesn't exceed 2⋅1052⋅105.
Output
For each test case, output a single integer denoting the minimum amount of minutes required for Victor to obtain both needed skills and −1−1 in case it's impossible to obtain the two skills after reading any amount of books.
Example
Input
Copy
6
4
2 00
3 10
4 01
4 00
5
3 01
3 01
5 01
2 10
9 10
1
5 11
3
9 11
8 01
7 10
6
4 01
6 01
7 01
8 00
9 01
1 00
4
8 00
9 10
9 11
8 11
Output
Copy
7
5
5
9
-1
8Note
In the first test case, we can use books 22 and 33, with a total amount of minutes spent equal to 3+4=73+4=7.
In the second test case, we can use the books 11 and 44, with a total amount of minutes spent equal to 3+2=53+2=5.
In the third test case, we have only one option and that is reading book 11 for a total amount of minutes spent equal to 55.
解题说明：此题是一道模拟题，可以采用贪心算法，找出包含1和10的耗费最小时间组合或者是直接包含11的最小时间，两者进行比较即可。
cpp
#include <stdio.h>
int main()
{
int t = 0, n = 0, a = 0, s = 0;
scanf("%d", &t);
while (t--)
{
scanf("%d", &n);
int min_1 = 200005, min_2 = 200005, min_3 = 2000005;
for (int i = 0; i < n; i++)
{
scanf("%d%d", &a, &s);
if (1 == s && min_1 > a)
{
min_1 = a;
}
else if (10 == s && min_2 > a)
{
min_2 = a;
}
else if (11 == s && min_3 > a)
{
min_3 = a;
}
}
if ((min_1 == 200005 || min_2 == 200005) && min_3 == 2000005)
{
printf("-1\n");
}
else if (min_1 + min_2 >= min_3)
{
printf("%d\n", min_3);
}
else
{
printf("%d\n", min_1 + min_2);
}
}
return 0;
}