ACM 01背包问题1

Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave … 
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ? 

 

Input

The first line contain a integer T , the number of cases. 
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

 

Output

One integer per line representing the maximum of the total value (this number will be less than 2 ³¹).

 

Sample Input

1

5 10

1 2 3 4 5

5 4 3 2 1

 

Sample Output

14

解题思路：

题目大意就是，给定一定数目的物体这些都有一定的体积和价值，需要我们将这些物体装到一个体积固定的背包中去，问我们需要怎样放才能使这个背包装的价值最大，需要我们输出这个最大的价值。这是一个01背包问题，我们对于一个物体只有放和不放这两种选择，套用01背包的模版可以解决这个问题.

程序代码：

#include<stdio.h>
#include<stdlib.h>
int  DP(int w[],int v[],int N,int M)
{
    int f[1024]={0};
   for( int i=1; i<=N; i++ )
   {
        for( int j=M; j>=0; j-- )
        {
             if( j>=v[i]&&f[ j ]<f[j-v[i]]+w[i] )
                 f[ j ]= f[ j-v[i] ] + w[i];   
        } 
   }
   return f[M];     
}
int  main()
{
    int n,N,M;
    int v[1005],w[1005];
    scanf( "%d",&n );
    for( int i=0; i<n; i++  )
    {
        scanf( "%d%d",&N,&M );
        for( int j=1; j<=N; j++ )
          scanf( "%d",&w[j] );
        for( int k=1; k<=N; k++ )
          scanf( "%d",&v[k] );
         printf( "%d\n",DP( w , v, N ,M ) );    
    }
    return 0;    
 
}