AC_4. I multiple knapsack problem

Code:

// multiple knapsack problem 
/ * 
F [i] of the total volume of the case of i, the maximum value is the number 
for (int i = 0; i <n-; i ++) 
{ 
    for (int J = V; J> = V [i ]; J,) 
    { 
        F [J] = max (F [J], F [JV [I]] + W [I], F [J-2 * V [I]] + 2 * W [I] ...) 
    } 
} 
1.f [I] = 0 
F [V] 
2.f [0] = 0, F [I] = INF, I! 0 = 
max (F [0 ... m]) 
* / 
#include <the iostream> 
#include <CString> 
#include <algorithm>
 the using  namespace STD;
 int N, V; // number of items and the capacity of the knapsack 
const  int NUM = 1005 ;
 int DP [NUM];
 int main()
{
    cin >> N >> V;
    for (int i = 0; i < N; i++)
    {
        int v, w, n;//输入每件物品的体积  价值 和 个数
        cin >> v >> w >> n;
        for (int j = V; j >= 0; j--)//枚举所有的体积
        {
            for (int k = 0; k <= n && k*v<=j; k++)//看看每个体积下的价值最大是多少
            {
                dp[j] = max(dp[j], dp[j - k*v] + k*w);//在计算的过程中每次存储最大价值
            }
        }
    }
    cout << dp[V] << endl;
    return 0;
}