HDU 1712 ACboy needs your help （分组背包）

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本文介绍了一种典型的分组背包问题，并提供了一个具体的实例来解释如何通过编程解决此类问题。通过对给定的课程和学习天数进行最优分配，实现最大化收益的目标。

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

大致题意：一个人有n门课，准备在m天内学习这些课，然后给你一个n*m的矩阵，a[i][j]表示第i门课学习j天的收益，一天只能学习一门课，问最大收益多少。

思路：比较裸的分组背包
代码如下

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
using namespace std; 
#define LL long long int
int a[105][105];
int dp[105];
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(!(n+m)) break;
        memset(dp,0,sizeof dp);
        for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        scanf("%d",&a[i][j]);
        for(int i=1;i<=n;i++)//i组
        for(int j=m;j>=0;j--)
        for(int k=1;k<=m;k++)
            if(j-k>=0)
            {
                dp[j]=max(dp[j],dp[j-k]+a[i][k]);
            }
        printf("%d\n",dp[m]);
    }

}