解题

#include <iostream>
#include "bit"
#include "vector"
#include "unordered_set"
#include "unordered_map"
#include "set"
#include "queue"
#include "algorithm"
#include "bitset"
#include "cstring"
#include "cmath"

using namespace std;

const int N = 5e4 + 1;
// 强度 >= P 的最小花费
int t, f[101][N], p[101], c[101]; // fij: 从前 i 个里选择，费用为 j 的最大强度

int main() {
cin >> t;
while (t--) {
int n, P, Q;
cin >> n >> P >> Q;
for (int i = 1; i <= n; i++) {
cin >> p[i] >> c[i];
}

    int x = 1e9;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= Q; j++) {
            f[i][j] = f[i - 1][j];
            if (j >= c[i])
                f[i][j] = max(f[i][j], f[i - 1][j - c[i]] + p[i]);
            if (f[i][j] >= P && j < x)
                x = j;
        }
    }
    if (x > Q) cout << "-1\n";
    else cout << x << endl;

}

}