hdu 1712 ACboy needs your help（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): 4185    Accepted Submission(s): 2220

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 21 21 32 22 12 12 33 2 13 2 10 0

 

Sample Output

346

 

  分组背包入门题

代码：

//15ms#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>using namespace std;const int maxn=100+10;int a[maxn][maxn];int dp[maxn];int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        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++)//根据不同课程分组        {            for(int v=m;v>=1;v--)//枚举不同容量            {                for(int j=1;j<=v;j++)//不同时间的课程                {                    dp[v]=max(dp[v],dp[v-j]+a[i][j]);                }            }        }        printf("%d\n",dp[m]);    }    return 0;}