Bone Collector(01背包）

Bone Collector

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 46 Accepted Submission(s) : 28

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

Author

Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

#include<iostream>#include<cstring>#include<algorithm>using namespace std;int main(int ac,char av[]){    int cases,n,bag_w,i,j,bone_w[1002],val[1002],dp[1002];    cin>>cases;    while(cases--)    {        cin>>n>>bag_w;        for(i=1;i<=n;i++)        cin>>val[i];        for(i=1;i<=n;i++)        cin>>bone_w[i];        memset(dp,0,sizeof(dp));        for(i=1;i<=n;i++)         for(j=bag_w;j>=bone_w[i];j--)         {             dp[j]=max(dp[j],dp[j-bone_w[i]]+val[i]);         }        cout<<dp[bag_w]<<endl;    }    return 0;}