HDU 2602 Bone Collector

E - Bone Collector

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status

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 
     
     
     
     


     
     
     
     

      
      一个简单的01背包
     
     
     
     

      
      #include<stdio.h>
#include<string.h>


#define MAX 100001


int dp[MAX],w[MAX],p[MAX];


int max(int a,int b)
{
    return a > b? a:b;
}
int main()
{
    int T,i,j;
    int n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&w[i]);
        }
        for(i=1;i<=n;i++)
        {
            scanf("%d",&p[i]);
        }
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
        {
            for(j=m;j>=p[i];j--)
            {
                dp[j] = max(dp[j], dp[j-p[i]]+w[i]);
            }
        }
        printf("%d\n",dp[m]);
    }
    return 0;
}