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  • ACboy needs your help (动态规划背包)

    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

    题解:动态规划背包问题,三重循环解即可

    代码:

    #include<cstdio>
    #include<iostream>
    #include<algorithm>
    #include<cstring>
    using namespace std;
    
    int main()
    {
        int n,m;
    	while(scanf("%d%d",&n,&m)!=EOF)
    	{
    	    if(n==0||m==0)
    		{
    		break;	
    		}	
    		long long int dp[105];
    		int a[105][105];
    		memset(dp,0,sizeof(dp));
            memset(a,0,sizeof(a));
    
    		for(int t=1;t<=n;t++)
    		{
    			for(int j=1;j<=m;j++)
    			{
    				scanf("%d",&a[t][j]);
    			}
    		}
    		for(int t=1;t<=n;t++)
    		{
    			for(int j=m;j>=0;j--)
    			{
    				for(int k=0;k<=j;k++)
    				{
    					dp[j]=max(dp[j],dp[j-k]+a[t][k]);
    				}
    			}
    		}
    		printf("%lld
    ",dp[m]);
    	}	
      return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Staceyacm/p/10782049.html
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