HD1712ACboy needs your help（纯裸分组背包）

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5589    Accepted Submission(s):
3043

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

 

题意：N个任务M天完成，每个任务花费每天都有一定的收益，问收益最大

分析：分组背包

把N个任务看成N组，其中每组中只能选择一个，也就是每一个任务花费的天数肯定是一个数，然后花费的天数还有费用，分组背包纯裸模板

http://www.cppblog.com/Onway/archive/2010/08/09/122695.html这个讲讲解了第二重和三重循环的设计思路

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 const int MAX = ;
 int n,m;
 int dp[MAX],a[MAX][MAX];
 int main()
 {
     while(scanf("%d%d", &n, &m) != EOF)
     {
         if(n ==  && m == )
             break;
         for(int i = ; i <= n; i++)
             for(int j = ; j <= m; j++)
                 scanf("%d", &a[i][j]);
         memset(dp, , sizeof(dp));
         for(int i = ; i <= n; i++)
         {
             for(int j = m; j > ; j--)
             {
                 for(int k = ; k <=m; k++)
                 {
                     if(j >= k)
                     dp[j] = max(dp[j], dp[j - k] + a[i][k]);
                 }
             }
         }
         printf("%d\n", dp[m]);
     }
     return ;
 }