HDU 1712 ACboy needs your help

分组背包问题，具体详情见背包九讲。、

ACboy needs your help

Time Limit : 1000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 9   Accepted Submission(s) : 6

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

   
   
   
   
   
   
   
   


   
   
   
   

#include <iostream>
#include <algorithm>
using namespace std;
int n,m;
int mp[105][105];
int dp[105];
int main()
{
	while( scanf("%d%d",&n,&m) == 2 ){
		if( n==0 && m==0 )
			break;
		for( int i=1 ; i<=n ; i++ )
			for( int j=1 ; j<=m ; j++ ) 
				scanf("%d",&mp[i][j]);
		memset( dp , 0 , sizeof(dp) );
		//注意这里的三层循环的顺序 “for v=V..0”这一层循环必须在
		//“for 所有的i属于组k”之外。
		//这样才能保证每一组内的物品最多只有一个会被添加到背包中。
		for( int i=1 ; i<=n ; i++ ){
			for( int V=m ; V>=0 ; V-- ){
				for( int k=1 ; k<=m ; k++ ){
					if( V-k>=0 )
						dp[V] = max( dp[V] , dp[V-k]+mp[i][k] );	
				}	
			}
		}
		//for( int i=0 ; i<=m ; i++ )
		printf("%d\n",dp[m]);
	}
	return 0;
}