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 = 110;
int n,m;
int dp[MAX],a[MAX][MAX];
int main()
{
    while(scanf("%d%d", &n, &m) != EOF)
    {
        if(n == 0 && m == 0)
            break;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                scanf("%d", &a[i][j]);
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i++)
        {
            for(int j = m; j > 0; j--)
            {
                for(int k = 1; k <=m; k++)
                {
                    if(j >= k)
                    dp[j] = max(dp[j], dp[j - k] + a[i][k]);
                }
            }
        }
        printf("%d\n", dp[m]);
    }
    return 0;
}