Nine Backpacks: Multiple Backpacks Problem I

tags: Nine Lectures on Backpack algorithm Dynamic programming c++

There are N items and a backpack with a capacity of V.

The i-th item has at most s_i Pieces, the volume of each piece is v_i, The value is w_i。

Solve which items are loaded into the backpack, so that the total volume of the items does not exceed the capacity of the backpack, and the total value is the largest.
Output the maximum value.

Input format
The two integers in the first line, N and V, are separated by a space, indicating the number of objects and the volume of the backpack respectively.

Next, there are N rows, each with three integers v_i,w_i,s_i, Separated by spaces, respectively indicate the volume, value and quantity of the i-th item.

Output format
outputs an integer that represents the maximum value.

data range
0<N,V≤100
0<v_i,w_i,s_i≤100
Input sample

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

Sample output:

10

#include <iostream>
using namespace std;
int fun[105][105];//fun[i][j] means that only the ith item can be selected from the previous i items, and the maximum value of the volume less than or equal to j
int main()
{
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++){// loop n items
        int v,w,s;
        cin>>v>>w>>s;//Enter item volume, value, quantity
        for(int j=m;j>=0;j--){//Loop volume, here cannot be j>=v, because if a two-dimensional array is stored, even if it is not selected, fun[i][j]=fun[i-1][j]
            for(int k=0;k<=s&&k*v<=j;k++){//Here when k=0 is to update fun[i][j]=fun[i-1][j] means that the i-th item is not selected
                fun[i][j]=max(fun[i][j],fun[i-1][j-k*v]+k*w);//Select k items i
            }
        }
    }
    cout<<fun[n][m]<<endl;
    return 0;
}

Binary optimized version

Binary optimized version of multiple knapsack problem

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