HDU【2602】Bone

Bone Collector

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

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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 
      
      
      
      
      
      
      
      

       
       
       
       #include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = 1005;


int v[maxn],w[maxn];
int f[maxn][maxn];


int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(f,0,sizeof(f));
        memset(v,0,sizeof(v));
        memset(w,0,sizeof(w));
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i = 1; i <= n; i++)
            scanf("%d",&v[i]);
        for(int i = 1; i <= n; i++)
            scanf("%d",&w[i]);
        for(int i = 1; i <= n; i++)
            for(int j = 0; j <= m; j++)
                if(j >= w[i])
                    f[i][j] = max(f[i-1][j] , f[i-1][j-w[i]] + v[i]);
                else f[i][j] = f[i-1][j];
        cout << f[n][m] << endl;
    }
    return 0;
}