HDU 1712 分组背包问题

题目信息：

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5146    Accepted Submission(s): 2780

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
  
  

题目大意：

一个学生用M天的时间复习N门课程，每门课程花费不同的天数，有不同的收获。问如何安排这M天，使得收获最大。

解题思路：

分组背包即物品分为多组，每一组的各个物品最多只能取一个。解题方法即在循环最里面再嵌套一个循环，查找每组中价值最大的物品。

代码部分：

#include <stdio.h>
#include <string.h>
int main()
{
	int N,M,i,j,k;
	int dp[105],value[105];
	while(scanf("%d %d",&N,&M)&&M||N)
	{
		memset(dp,0,sizeof(dp));
		memset(value,0,sizeof(value));
		for(i=1;i<=N;i++)
		{
			for(j=1;j<=M;j++)
			{
				scanf("%d",&value[j]);
			}
			for(j=M;j>0;j--)
			{
				for(k=j;k>0;k--)
					dp[j]=dp[j]>dp[j-k]+value[k]?dp[j]:dp[j-k]+value[k];
			}
		}
		printf("%d\n",dp[M]);
	}
	return 0;
}