ACboy needs your help hdu 分组背包问题

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：  有n个课程，m天时间，接下来n*m的矩阵表示该课程花费1~m天时间所获能得的价值，求花费m天时间能获得的最大价值
   背包：
       分组背包模板题..将每一天看成一个组，每组只能选一个。

代码实现：

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cstring>
 5 using namespace std;
 6 int max(int a,int b)
 7 {
 8     return a>b?a:b;
 9 }
10 int main()
11 {
12     int n,m,i,j,k,a[105][105],dp[105];
13     while(cin>>n>>m&&n&&m)
14     {
15         for(i=1;i<=n;i++)
16         for(j=1;j<=m;j++)
17         cin>>a[i][j];
18         memset(dp,0,sizeof(dp));
19         for(i=1;i<=n;i++)//分组
20         for(j=m;j>0;j--)//容量
21         for(k=1;k<=j;k++)//第i组数据
22         dp[j]=max(dp[j],dp[j-k]+a[i][k]);
23         cout<<dp[m]<<endl;
24     }
25     return 0;
26 }