HDUOJ---1712 ACboy needs your help

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3282    Accepted Submission(s): 1703

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！....

/*o1背包@龚细军*/
/*维度为2的01背包*/
#include<stdio.h>
#include<string.h>
#define maxn 102
int dp[maxn];
int aa[maxn][maxn];

int max(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    int m,n,i,j,k;
    while(scanf("%d%d",&n,&m),m+n)
    {
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)        // class
            for(j=1;j<=m;j++)    //day
                scanf("%d",&aa[i][j]);
            //对每一门课程进行背包施放
            for(i=1;i<=n;i++)
            {
               for(j=m;j>=0;j--)  //代表的是背包的容量
               {
                   for(k=1;k<=j;k++)  //在这个容量内选择一个房间去，和之前放进去的比较！
                       dp[j]=max(dp[j],dp[j-k]+aa[i][k]);
               }
            }
         printf("%d
",dp[m]);
    }
    return 0;
}