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 2³¹).

 

Sample Input

15 101 2 3 4 55 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 1001int 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;}