HDU ACboy needs your help (分组背包)

ACboy needs your help

Time Limit : 1000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 2   Accepted Submission(s) : 2

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

 

 

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

int n,m;
int a[110][110];
int dp[110];

int main(){

    //freopen("input.txt","r",stdin);

    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++)   //第一重循环：分组数
            for(int j=m;j>=0;j--)   //第二重循环：容量体积
                for(int k=1;k<=j;k++)   //第三重循环：属于i组的k
                    dp[j]=max(dp[j],dp[j-k]+a[i][k]);
        printf("%d\n",dp[m]);
    }
    return 0;
}