ACboy needs your help（分组背包）

比眉伴天荒 2022-08-01 00:07 101阅读 0赞

# ACboy needs your help #

**Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 4867    Accepted Submission(s): 2621**

Problem Description

ACboy has N courses this term, and he plans to spend at most M days on study.Of course,the profit he will gain from different course depending on the days he spend on it.How to arrange the M days for the N courses to maximize the profit?

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers N and M, N is the number of courses, M is the days ACboy has.  
Next follow a matrix A\[i\]\[j\], (1<=i<=N<=100,1<=j<=M<=100).A\[i\]\[j\] indicates if ACboy spend j days on ith course he will get profit of value A\[i\]\[j\].  
N = 0 and M = 0 ends the input.

Output

For each data set, your program should output a line which contains the number of the max profit ACboy will gain.

Sample Input

2 2
    1 2
    1 3
    2 2
    2 1
    2 1
    2 3
    3 2 1
    3 2 1
    0 0

Sample Output

3
    4
    6

** **

**分组背包问题：**

** 有N件物品和一个容量为V的背包。第i件物品的费用是c\[i\]，价**

**值是w\[i\]。这些物品被划分为若干组，每组中的物品互相冲突，**

**最多选一件。求解将哪些物品装入背包可使这些物品的费用总和**

**不超过背包容量，且价值总和最大。**

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<iostream>
    const int MAX=105;
    using namespace std;
    int dp[MAX];
    struct node
    {
        int v;
        int w;
    }N[MAX][MAX];
    int main()
    {
        int n,m,i,j,k;
        while(cin>>n>>m,(n+m))
        {
            memset(dp,0,sizeof(dp));
            memset(N,0,sizeof(N));
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=m;j++)
                {
                    cin>>N[i][j].v;
                    N[i][j].w=j;
                }
            }
            for(i=1;i<=n;i++)
            {
                for(j=m;j>=0;j--)
                {
                    for(k=1;k<=m;k++)
                    {
                        if(j>=N[i][k].w)
                        {
                            dp[j]=max(dp[j],dp[j-N[i][k].w]+N[i][k].v);
                        }
                    }
                }
            }
            cout<<dp[m]<<endl;
        }
    }