DP 集合定义：类似01背包，完全背包
DP[i][j]:选完第i个物品后，容量是j的背包，(选法)方案的最大价值
f[i][j]=max(f[i][j],f[i-1][j-kv[i]]+kw[i]);

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 1010;
int main()
{
    int n, V, f[N][N],v[N],w[N],s[N];
    cin >> n >> V;
    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 <=V; j ++ ){
            for (int k = 0; k <= s[i] && k*v[i]<=j; k ++ ){
                    f[i][j] = max(f[i][j],f[i -1][j-k*v[i]]+k*w[i]);
            }
        }
    }
    cout << f[n][V] << endl;
    return 0;
}

优化：
等价变形：

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 1010;
int main()
{
    int n, V, f[N],v[N],w[N],s[N];
    cin >> n >> V;
    for (int i = 1; i <= n; i ++ ){
        cin >> v[i] >> w[i] >> s[i];
    }
    for (int i = 1; i <= n; i ++ ){
        for (int j = V; j >=0; j -- ){ //和01背包优化一个原理。递推顺序，反着的话f[i],f[i-1]顺序是对的，还是上一轮的f[i-1]而不是当前i轮更新过的
            for (int k = 0; k <= s[i] && k*v[i]<=j; k ++ ){
                    f[j] = max(f[j],f[j-k*v[i]]+k*w[i]);
            }
        }
    }
    cout << f[V] << endl;
    return 0;
}