HDU2602-Bone Collector

Bone Collector

                                                                             Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                      Total Submission(s): 52312    Accepted Submission(s): 22049

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
  

 

Author

Teddy

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

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

using namespace std;

#define N 1001
int dp[N];
int c[N],w[N];

int main()
{
    int t,n,v;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&v);
        for(int i=0;i<n;i++)
            scanf("%d",&c[i]);
        for(int i=0;i<n;i++)
            scanf("%d",&w[i]);
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)
        {
            for(int j=v;j>=w[i];j--)
                dp[j]=max(dp[j],dp[j-w[i]]+c[i]);
        }
        printf("%d\n",dp[v]);
    }
    return 0;
}