ACboy needs your help（HDU 1712 分组背包入门）

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5872 Accepted Submission(s): 3196

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

Source

HDU 2007-Spring Programming Contest

分组背包，背包容量m,物品分为n组，每组只能取一件，求背包最大价值。

dp[i][j]表示对于前i组物品，背包容量为j时的最大价值，此时对于每种dp[i][j]需要遍历第i组的每一个物品,求出最大的dp[i][j];

状态转移方程：

　　　　　　　　　　dp[i][j]=max(dp[i][j],dp[i-1][j-k]+ma[i][k])

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <algorithm>
 5 using namespace std;
 6 #define Max 105
 7 int dp[Max][Max],ma[Max][Max];
 8 int n,m;
 9 int main()
10 {
11     int i,j;
12     memset(ma,0,sizeof(ma));
13     freopen("in.txt","r",stdin);
14     while(scanf("%d%d",&n,&m))
15     {
16         if(n==0&&m==0)
17             break;    
18         for(i=1;i<=n;i++)
19             for(j=1;j<=m;j++)
20                 scanf("%d",&ma[i][j]);
21         memset(dp,0,sizeof(dp));
22         for(i=1;i<=n;i++)
23         {
24             for(j=1;j<=m;j++)
25             {
26                 for(int k=0;k<=j;k++)
27                 {
28                     if(dp[i][j]<dp[i-1][j-k]+ma[i][k])
29                         dp[i][j]=dp[i-1][j-k]+ma[i][k];
30                 }
31                 //cout<<dp[i][j]<<" ";
32             }
33         //    cout<<endl;
34         }
35         printf("%d
",dp[n][m]);
36     }
37     return 0;
38 }