hdu1712 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): 1768    Accepted Submission(s): 890

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

 

分析：裸的分组背包（详解可以看看背包九讲）

#include<iostream>
#include<cstdio>
#define MAXV 110
#define MAXN 110

using namespace std;

int d[MAXV],a[MAXN][MAXN];

int main()
{
    int n,m,i,j,k;
    while(scanf("%d%d",&n,&m)!=EOF&&(n||m))
    {
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
                scanf("%d",&a[i][j]);
        for(i=0;i<=m;i++)
            d[i]=0;
        for(i=1;i<=n;i++)
        {
            for(j=m;j>0;j--)
            {
                for(k=1;k<=m;k++)
                {
                    if(j-k>=0)
                        d[j]=max(d[j],d[j-k]+a[i][k]);
                }
            }
        }
        printf("%d\n",d[m]);
    }
    return 0;
}