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): 6953 Accepted Submission(s): 3853

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

题意：
n门课程m天时间，每天最多选一门课程，A [i] [j]表示花费j天，获得价值A [i] [j]的价值，如何安排使得价值最大。
题解：分组01背包，分成n组，容量为m，价值为a[i][j],每组选取一门课程花费k天，获得最大价值。

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int maxn = 105;int n,m;int a[maxn][maxn];int dp[maxn];int main(){    while(cin>>n>>m&&(m+n))    {        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)            {                cin>>a[i][j];            }        }        for(int i=1;i<=n;i++)//n组        {            for(int j=m;j>=1;j--)//天数从小到大            {                 for(int k=j;k>=1;k--)//01                {                    dp[j] = max(dp[j],dp[j-k]+a[i][k]);                }            }        }        cout<<dp[m]<<endl;        memset(dp,0,sizeof(dp));    }    return 0;}