hdu 1712 ACboy needs your help 分组背包

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1712

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

题意：一開始输入n和m，n代表有n门课，m代表你有m天，
然后给你一个数组，val[i][j]，代表第i门课，在通过j天去修，
会得到的分数。求在m天能得到的最大分数。

#include <cstdio>
#include <cstring>
#define N 147
int max(int a, int b)
{
	if(a > b)
		return a;
	return b;
}
int main()
{
	int n,m,a[N][N],dp[N];
	int i, j, k;
	while(~scanf("%d%d",&n,&m)&&( n || m))
	{
		memset(dp,0,sizeof(dp));
		for(i = 1; i <=n ; i++)
		{
			for(j = 1; j <= m ;j++)
			scanf("%d",&a[i][j]);
		}
		for(i = 1 ; i <= n ; i++)//第一重循环：分组数
		{
			for(j = m ; j >= 1 ; j--) //第二重循环：容量体积
			{
				for(k = 1 ; k <= j ; k++) //第三重循环：属于i组的k
				{
					dp[j]=max(dp[j],dp[j-k]+a[i][k]);
				}
			}
		}
		printf("%d
",dp[m]);
	}
	return 0;
}