HDU 2602 Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 59414    Accepted Submission(s): 24807

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<stdio.h>#include<math.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;struct p{    int x,y;///x,y分别表示价值和体积}f[1005];int main(){    int t;    int dp[1005];    cin>>t;    while(t--)    {        memset(dp,0,sizeof(dp));///初始化        int n,v;            cin>>n>>v;        for(int i=0;i<n;i++)            cin>>f[i].x;        for(int i=0;i<n;i++)            cin>>f[i].y;        for(int i=0;i<n;i++)            for(int j=v;j>=f[i].y;j--)            dp[j]=max(dp[j],dp[j-f[i].y]+f[i].x);        cout<<dp[v]<<endl;    }    return 0;}