hdu 2602--Bone Collector（01背包）

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

 

01背包：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <iostream>
using namespace std;
int dp[10000];

int main()
{
    int t,i,j,n,v;
    scanf("%d",&t);
    while(t--){
        int a[10000];
        int b[10000];
        memset(dp,0,sizeof(dp));
        scanf("%d%d",&n,&v);
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(i=1;i<=n;i++)
            scanf("%d",&b[i]);
        for(i=1;i<=n;i++){
            for(j=v;j>=b[i];j--){
                dp[j]=max(dp[j],dp[j-b[i]]+a[i]);
            }
        }
        printf("%d
",dp[v]);
    }
    return 0;
}