The knapsack problem of two-dimensional cost

Yes NN Items and a capacity are VV The backpack , The maximum weight a backpack can bear is MM.

Each item can only be used once . The volume is vivi, The weight is mimi, The value is wiwi.

Find out which items to pack in your backpack , It can make the total volume of the goods not exceed the capacity of the backpack , The total weight shall not exceed the maximum weight that the backpack can bear , And the sum of values is the largest .
Output maximum value .

Input format

The first row has three integers ,N,V,MN,V,M, Space off , They represent the number of items 、 Backpack volume and maximum weight that backpack can bear .

Next there is NN That's ok , Three integers per row vi,mi,wivi,mi,wi, Space off , Separate indication control ii The volume of items 、 Weight and value .

Output format

Output an integer , Represents the greatest value .

Data range

0<N≤10000<N≤1000
0<V,M≤1000<V,M≤100
0<vi,mi≤1000<vi,mi≤100
0<wi≤10000<wi≤1000

sample input

4 5 6
1 2 3
2 4 4
3 4 5
4 5 6

sample output ：

8

#include <bits/stdc++.h>

using namespace std;

const int N = 110;

int n, V, M;
int f[N][N];

int main()
{
    scanf("%d%d%d", &n, &V, &M);
    
    for(int i = 1; i <= n; i ++)//01 Backpack first dimension 
    {
        int v, m, w;
        scanf("%d%d%d", &v, &m, &w);
        for(int j = V; j >= v; j --)//01 Backpack 2D 
            for(int k = M; k >= m; k --)//01 Backpack 2D 
                f[j][k] = max(f[j][k], f[j - v][k - m] + w);
    }
    
    printf("%d", f[V][M]);
    
    return 0;
}