hdu 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): 2793    Accepted Submission(s): 1441 

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

 

Source

HDU 2007-Spring Programming Contest

 

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lcy

// 1A 看来我的想法还是没有问题的、、
// 因为同一类不能叠加更新，所以选取了 2个数组 ，然后就是 背包了
#include <iostream>
#include <algorithm>
#include <queue>
#include <math.h>
#include <stdio.h>
#include <string.h>
using namespace std;
int A[110][110];
int dp1[110],dp2[110];
int main()
{
    int n,m;
    int in;
    int i,j,k;
    while(scanf("%d %d",&n,&m),n|m){

       for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
         scanf("%d",&A[i][j]);
      memset(dp1,0,sizeof(dp1));
      memset(dp2,0,sizeof(dp2));
      for(i=1;i<=n;i++){
        for(j=1;j<=m;j++)
         for(k=m;k>=j;k--){
            if(dp2[k]<dp1[k-j]+A[i][j])
                dp2[k]=dp1[k-j]+A[i][j];
         }
         for(j=1;j<=m;j++)
            dp1[j]=dp2[j];
      }
      printf("%d
",dp1[m]);


    }

    return 0;
}