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

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 842    Accepted Submission(s): 419

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

 

Recommend

lcy

 

题目： http://acm.hdu.edu.cn/showproblem.php?pid=1712

分析：这题是裸的分组背包，分组背包最重要的地方就是保证每组里面的物品不同时取，这个可以通过改变枚举物品顺序做到，分三重枚举，第一重枚举第几组，第二重枚举背包容量，要倒着来，第三重枚举该组里面的物品。。。详细请看背包九讲

代码：

#include<cstdio>
#include<iostream>
using namespace std;
const int mm=111;
int f[mm],a[mm][mm];
int i,j,k,n,m;
int main()
{
    while(scanf("%d%d",&n,&m),n+m)
    {
        for(i=0;i<=m;++i)f[i]=0;
        for(i=0;i<n;++i)
            for(j=1;j<=m;++j)scanf("%d",&a[i][j]);
        for(i=0;i<n;++i)
            for(j=m;j>=0;--j)
                for(k=1;k<=j;++k)
                    f[j]=max(f[j],f[j-k]+a[i][k]);
        printf("%d\n",f[m]);
    }
    return 0;
}