HDU-1712-ACboy needs your help（分组背包）

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 gainfrom different course depending on the days he spend on it.How to arrange the Mdays for the N courses to maximize the profit? 

Input

The input consists of multipledata sets. A data set starts with a line containing two positive integers N andM, 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 spendj 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, yourprogram should output a line which contains the number of the max profit ACboywill 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

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题意：有n门课，你有m天时间。一天只能上一门。一门课上一天的利益可能比上两天还多，但可能比三天少。也就是说，每门课只能得到上的天数的相应利益，而不是累加利益。问最大利益是多少。

思路：分组背包问题。

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//分组背包 
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int map[105][105];//i课程花j天的利润 
int dp[105];//花x天时的利润 
int n,m;

int main(){
    while(scanf("%d%d",&n,&m)){
        if(n==0&m==0) break;
        memset(map,0,sizeof(map));
        memset(dp,0,sizeof(dp));
        
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                scanf("%d",&map[i][j]);
            }
        }

        for(int i=1;i<=n;i++){
            for(int j=m;j>=1;j--){//j是容量，每找一个课程更新一下最大利润 
                for(int k=1;k<=j;k++){//k是所耗天数， 
                    dp[j]=max(dp[j],dp[j-k]+map[i][k]);//j-k耗k天之前的利润 
                }
            }
        }
        printf("%d\n",dp[m]);         
    }    
    return 0;
}