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

3

4

6

 

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])

 

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <algorithm>
 using namespace std;
 #define Max 105
 int dp[Max][Max],ma[Max][Max];
 int n,m;
 int main()
 {
     int i,j;
     memset(ma,,sizeof(ma));
     freopen("in.txt","r",stdin);
     while(scanf("%d%d",&n,&m))
     {
         if(n==&&m==)
             break;
         for(i=;i<=n;i++)
             for(j=;j<=m;j++)
                 scanf("%d",&ma[i][j]);
         memset(dp,,sizeof(dp));
         for(i=;i<=n;i++)
         {
             for(j=;j<=m;j++)
             {
                 for(int k=;k<=j;k++)
                 {
                     if(dp[i][j]<dp[i-][j-k]+ma[i][k])
                         dp[i][j]=dp[i-][j-k]+ma[i][k];
                 }
                 //cout<<dp[i][j]<<" ";
             }
         //    cout<<endl;
         }
         printf("%d\n",dp[n][m]);
     }
     return ;
 }