HDU 1712 ACboy needs your help（分组背包）

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ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6524 Accepted Submission(s): 3608

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 21 21 32 22 12 12 33 2 13 2 10 0

Sample Output

Source

HDU 2007-Spring Programming Contest

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题目大意：
有一个小孩有N门课程要复习，对于每门课程花费不同的天数会得到不同的收益，现在他只有M天，问他可以达到的最大收益。

解题思路：
    分组背包入门题。可以把每门课程看成一组，每一个课程的不同时间看成不同的物品。然后直接分组背包即可。
    附上背包九讲的讲解：

AC代码：

#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <algorithm>using namespace std;const int MAXN=100+3;int N,M,val[MAXN][MAXN],dp[MAXN];int main(){    while(~scanf("%d%d",&N,&M))    {        if(!N&&!M)            break;        for(int i=1;i<=M;++i)//初始化            dp[i]=0;        for(int i=1;i<=N;++i)            for(int j=1;j<=M;++j)                scanf("%d",&val[i][j]);        for(int k=1;k<=N;++k)//N组            for(int v=M;v>=0;--v)//使用的天数                for(int i=0;i<=v;++i)//遍历每一组的所有物品                    dp[v]=max(dp[v],dp[v-i]+val[k][i]);        int ans=0;        for(int i=0;i<=M;++i)            ans=max(ans,dp[i]);        printf("%d\n",ans);    }        return 0;}

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