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 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

  
  

 

分组背包问题，详见背包九讲。

分析：将天数m作为背包的容量，科目数目作为背包的种类数目，天数j作为背包的重量，因为一个科目只能选一次，对应于

每组中的物品只能选一件，正好是分组背包问题。

方程：dp[j]=max(dp[j-k]+a[i][k],dp[j])

代码：

#include<stdio.h>
#include<string.h>
int dp[105],a[105][105];
int max(int a,int b)
{
       if(a<b)
           return b;
       return a;
}
int main()
{
       int i,j,k,n,m;
       while(scanf("%d%d",&n,&m)!=EOF)
       {
              if(n==0&&m==0)
                  break;
              memset(dp,0,sizeof(dp));
              for(i=0;i<n;i++)
                for(j=1;j<=m;j++)
                  scanf("%d",&a[i][j]);
              for(i=0;i<n;i++)
                  for(j=m;j>=0;j--)
                     for(k=1;k<=j;k++)
                       dp[j]=max(dp[j],dp[j-k]+a[i][k]);
               printf("%d\n",dp[m]);
       }
       return 0;
}