HDOJ 1712 ACboy needs your help

分组背包

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2665    Accepted Submission(s): 1359

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[j], (1<=i<=N<=100,1<=j<=M<=100).A[j] indicates if ACboy spend j days on ith course he will get profit of value A[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

 

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

int a[111][111];

int dp[2][111];

int n,m;

int main()

{

while(scanf("%d%d",&n,&m))

{

    if(n==0&&m==0) break;

    memset(a,0,sizeof(a));

    memset(dp,0,sizeof(dp));

    for(int i=1;i<=n;i++)

    {

        for(int j=1;j<=m;j++)

            scanf("%d",&a[j]);

    }

    for(int k=1;k<=n;k++)

    {

        for(int i=1;i<=m;i++)

        {

           for(int j=i;j>=0;j--)

           {

               dp[k&1]=max(dp[(k-1)&1][i-j]+a[k][j],dp[k&1]);

           }

        }

    }

    int ans=-1;

    int t=n&1;

    for(int i=0;i<=m;i++)

        ans=max(ans,dp[t]);

    cout<<ans<<endl;

}

    return 0;

}