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

这题是分组背包，对于每一门课，有m种天数的选择。并且选一天和选两天是相互排斥的。就是说要么不选，要么选k天，所以就在01背包基础上多加一个循环，注意次序，m循环要在k之前，这样才不会发生一门科目反复选多天的情况。
#include<stdio.h>
#include<string.h>
int max(int a,int b){
	return a>b?
a:b;
}
int v[106][106],dp[106];
int main()
{
	int n,m,i,j,k;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		if(n==m && n==0)break;
		for(i=1;i<=n;i++){
			for(j=1;j<=m;j++){
				scanf("%d",&v[i][j]);
			}
		}
		memset(dp,0,sizeof(dp));
		for(i=1;i<=n;i++){
			for(j=m;j>=1;j--){
				for(k=1;k<=m;k++){
					if(j>=k && dp[j]<dp[j-k]+v[i][k]){
						dp[j]=dp[j-k]+v[i][k];
					}
				}
			}
		}
		printf("%d\n",dp[m]);
	}
	return 0;
}