思路

• 集合：所有只从前i个物品中选，且总体积不超过j的选法
• 状态计算：
  • 根据第i个物品选的个数划分集合
  • f[i][j] = max(f[i - 1][j - v[i] * k) + w[i] * k); k = 0, 1, 2, 3…

#include <iostream>

using namespace std;

const int N = 110;

int n, m;
int v[N], w[N], s[N];
int f[N][N];

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

    cout << f[n][m] << endl;

    return 0;
}