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Bone Collector dp 01背包

热度:51   发布时间:2023-11-18 06:29:01.0

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;
}