解题思路

朴素的多重背包，再朴素的完全背包的基础，加一个次数限制，即可
详情请看完全背包

代码

#include <iostream>

using namespace std;

const int N = 110;
int v[N], w[N], s[N], f[N][N];

int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> v[i] >> w[i] >> s[i];

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            for (int k = 0; j >= k * v[i] && k <= s[i]; k++) {
                f[i][j] = max(f[i][j], f[i - 1][j - k * v[i]] + k * w[i]);
            }
        }
    }
    cout << f[n][m] << endl;

    return 0;
}