HDU2602 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

 

Author

Teddy

代码：

#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn=1005;
int w[maxn],v[maxn],dp[maxn];

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n,V;
        scanf("%d%d",&n,&V);
        for(int i=1;i<=n;i++) scanf("%d",&w[i]);
        for(int i=1;i<=n;i++) scanf("%d",&v[i]);
        for(int i=0;i<=V;i++) dp[i]=0;
        for(int i=1;i<=n;i++)
            for(int j=V;j>=v[i];j--)
                dp[j]=max(dp[j],dp[j-v[i]]+w[i]);
        printf("%d\n",dp[V]);
    }
    return 0;
}