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): 5516    Accepted Submission(s): 3004

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

 

思路：分组背包

深刻理解分组背包的三个循环

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[101][101],dp[101];
int main()
{
    int day,sub;
    while(~scanf("%d %d",&sub,&day)&&(sub&&day))
     {
         memset(a,0,sizeof(a));
         for(int i=1;i<=sub;i++)
            for(int j=1;j<=day;j++)
                cin>>a[i][j];
         memset(dp,0,sizeof(dp));
         for(int i=1;i<=sub;i++)  //
            for(int j=day;j>=0;j--)   //对总的使用天数枚举dp[j]表示已使用完  j  天时的最大价值

                                                  //一定要逆序，因为后面dp[j-k]+a[i][k]可以发现后面的状态会使用到前面的初始                                                           //状态
             for(int k=0;k<=j;k++)  //该处是对第i组的物品（花费天数）进行枚举，第i组花费天数<=总花费天数
         dp[j]=max(dp[j],dp[j-k]+a[i][k]);   //最优子结构，花费j天的条件下所能得到的最大价值
         cout<<dp[day]<<endl;
     }
     return 0;
}