杭电 2602 Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 27413    Accepted Submission(s): 11154

Problem 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
   
     

01背包。

AC代码例如以下：

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    int t,n,v;
    int i,j;
    int a[1005],b[1005],dp[1005];
    cin>>t;
    while(t--)
    {
        cin>>n>>v;
        for(i=1;i<=n;i++)
            cin>>a[i];
        for(i=1;i<=n;i++)
            cin>>b[i];
        memset(dp,0,sizeof dp);
        for(i=1;i<=n;i++)
            for(j=v;j>=b[i];j--)
            dp[j]=max(dp[j-b[i]]+a[i],dp[j]);
            cout<<dp[v]<<endl;
    }

    return 0;
}