hdu2602

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25116    Accepted Submission(s): 10186

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

 

很基础的一道01背包

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
    int va[1010],vo[1010],dp[1010],i,j,T,n,v;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&v);
        memset(dp,0,sizeof(dp));
        for(i=0; i<n; i++)
            scanf("%d",&va[i]);
        for(i=0; i<n; i++)
            scanf("%d",&vo[i]);
        for(i=0; i<n; i++)
            for(j=v; j>=vo[i]; j--)
                dp[j]=max(dp[j],dp[j-vo[i]]+va[i]);
        printf("%d
",dp[v]);
    }
    return 0;
}