Problem:
Knapsack Problem with Limitations

You have N items that you want to put them into a knapsack. Item i has value vi, weight wi and limitation mi.

You want to find a subset of items to put such that:

The total value of the items is as large as possible.

The items have combined weight at most W, that is capacity of the knapsack.

You can select at most mi items for ith item.

Find the maximum total value of items in the knapsack.

Input

N W
v1 w1 m1
v2 w2 m2
:
vN wN mN

The first line consists of the integers N and W. In the following N lines, the value, weight and limitation of the i-th item are given.

Output

Print the maximum total values of the items in a line.

Constraints

1 ≤ N ≤ 100

1 ≤ vi ≤ 1,000

1 ≤ wi ≤ 1,000

1 ≤ mi ≤ 10,000

1 ≤ W ≤ 10,000

Sample Input 1

4 8
4 3 2
2 1 1
1 2 4
3 2 2

Sample Output 1

12

Sample Input 2

2 100
1 1 100
2 1 50

Sample Output 2

150

Solution in C++ (judge verdict: Accepted)

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<stack>
#include<climits>
#include<cstring>
#include<queue>
using namespace std;

const int INF = INT_MAX/3;

void slidingMinimum(vector<int> &inputList, vector<int> &outputList, int slideSize){
    deque<int> dq;
    for(int i=0;i<inputList.size();i++){
        while(!dq.empty()){
            if (inputList[dq.back()] >= inputList[i]){
                dq.pop_back();
            }else{
                break;
            }
        }
        dq.push_back(i);

        outputList.push_back(inputList[dq.front()]);
        if(dq.front() == i - slideSize +1){
            dq.pop_front();
        }
    }
    return;
}

int knapsackProblemWithLimitations(int W, vector<int> &v, vector<int> &w, vector<int> &m){
    int N = v.size();
    int DP[W+1][N];

    for (int i=0;i<W+1;i++){
        for (int j=0;j<N;j++){
            DP[i][j] = 0;
        }
    }

    for (int i=1;i< min(m[0], W/w[0])+1;i++){
        DP[w[0]*i][0] = v[0] * i;
    }

    for (int j=1;j<N;j++){
        for (int t=0;t<w[j];t++){
            vector<int> b,c;
            for (int k=0;(w[j]*k + t)<W+1;k++){
                b.push_back(DP[w[j]*k+t][j-1] - k*v[j]);
            }

            //??¢??°???????°????????±???????????????§???????±???????????????§??\???
            for (int i=0;i<b.size();i++){
                b[i] = - b[i];
            }
            slidingMinimum(b,c,m[j]+1);
            for (int i=0;i<c.size();i++){
                c[i] = - c[i];
            }
/*
            cout << endl;
            for (int p=0;p<b.size();p++){
                cout << b[p] << " ";
            }
            cout << endl <<endl;

            cout << endl;
            for (int p=0;p<c.size();p++){
                cout << c[p] << " ";
            }
            cout << endl <<endl;
*/
            for(int k=0;k<c.size();k++){
                DP[w[j]*k+t][j] = c[k] + k*v[j];
            }
        }
    }
/*
    cout << endl;
    for (int i=0;i<W+1;i++){
        for (int j=0;j<N;j++){
            cout << DP[i][j] << " ";
        }
        cout << endl;
    }
    cout << endl;
*/
    int S = 0;
    for (int i=0;i<W+1;i++){
        S = max(S, DP[i][N-1]);
    }
    return S;
}

int main(){
    int N, W;
    cin >> N >> W;
    vector<int> v(N);
    vector<int> w(N);
    vector<int> m(N);
    for (int i=0;i<N;i++){
        cin >> v[i] >> w[i] >> m[i];
    }

    cout << knapsackProblemWithLimitations(W, v, w, m) << endl;
    return 0;
}