Hdu oj 2602 Bone Collector（01背包）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 56560    Accepted Submission(s): 23661

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
#include<cstdio>#include<iostream>#include<cstring>using namespace std;const int Maxn = 1005;int v[Maxn];int w[Maxn];int dp[Maxn];int N,V;int t;int main(){    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&N,&V);        for(int i=1;i<=N;i++)        {            scanf("%d",&v[i]);        }        for(int i=1;i<=N;i++)        {            scanf("%d",&w[i]);        }        memset(dp,0,sizeof(dp));        for(int i=1;i<=N;i++)        {            for(int j=V;j>=w[i];j--)            {                dp[j] = max(dp[j],dp[j-w[i]] + v[i]);            }        }        printf("%d\n",dp[V]);    }}