状态表示：f[i][j] 从前i个物品中选出容量不超过j的物品的最大值

状态计算：f[i][j] = max(f[i][j], f[i-1][j-v]+w, f[i-1][j-2v]+2w, f[i-1][j-3v]+3w, …)

ps. 多重背包：好像没有优化的。

C++ 代码

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 110;

int v[N], w[N], res[N];
int f[N][N];
int main()
{
    int n, m;
    cin >> n >> m;

    for(int i = 1; i <= n; i++) cin >> v[i] >> w[i] >> res[i];

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

    cout << f[n][m];

    return 0;
}