Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 56560    Accepted Submission(s): 23661


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 1 5 10 1 2 3 4 5 5 4 3 2 1
Sample Output 14

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;

const int Maxn = 1005;
int v[Maxn];
int w[Maxn];
int dp[Maxn];
int N,V;
int t;

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&N,&V);
        for(int i=1;i<=N;i++)
        {
            scanf("%d",&v[i]);
        }
        for(int i=1;i<=N;i++)
        {
            scanf("%d",&w[i]);
        }

        memset(dp,0,sizeof(dp));

        for(int i=1;i<=N;i++)
        {
            for(int j=V;j>=w[i];j--)
            {
                dp[j] = max(dp[j],dp[j-w[i]] + v[i]);
            }
        }
        printf("%d\n",dp[V]);
    }
}