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

 

Recommend

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\n",dp1[m]);





    }



    return 0;

}