## Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25800    Accepted Submission(s): 10453

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 231).

Sample Input

15 101 2 3 4 55 4 3 2 1

Sample Output

14

#include<iostream>using namespace std;int max(int a,int b){if(a>b)return a;else return b;}int n,j,v,c[1004],w[1004],dp[1004][1004],k,i;int main(){  cin>>k;  while(k--)  {    memset(dp,0,sizeof(dp));    cin>>n>>v;    for(i=1;i<=n;i++)      cin>>c[i];    for(i=1;i<=n;i++)      cin>>w[i];    for(i=1;i<=n;i++)      for(j=0;j<=v;j++)        if(w[i]<=j)        dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+c[i]);        else                     dp[i][j]=dp[i-1][j];                     cout<<dp[n][v]<<endl;  }return 0;}