Bone Collector dp 01背包

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 65125    Accepted Submission(s): 27135


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

 

Source

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

用01背包一带就过了

#include<iostream>

#include<string.h>

#include<algorithm>

using namespace std;

#define Size 1111

int va[Size],vo[Size];

int dp[Size];

int Max(int x,int y)

{

    return x>y?x:y;

}

int main()

{

    int t,n,v;

    int i,j;

    cin>>t;

    while(t--)

    {

        cin>>n>>v;

        for(i=1;i<=n;i++)

            cin>>va[i];

        for(i=1;i<=n;i++)

            cin>>vo[i];

        memset(dp,0,sizeof(dp));

        for(i=1;i<=n;i++)

        {

            for(j=v;j>=vo[i];j--)

            {

                dp[j]=Max(dp[j],dp[j-vo[i]]+va[i]);

            }

        }

        cout<<dp[v]<<endl;

    }

    return 0;

}