HDOJ 2602 Bone Collector【01背包】

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 34251    Accepted Submission(s): 14101

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 2 ³¹).

 

Sample Input

   
   
   
   

    
    
    
    1
5 10
1 2 3 4 5
5 4 3 2 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    14
   
   
   
   

 

题解:01背包

AC代码:

#include<iostream>
#include<cstring>
#define N 10005
using namespace std;
struct{
    int v,c;
}Bag[N];
int n,v,f[N],t;
int main()
{
    cin.sync_with_stdio(false);
    cin>>t;
    while(t--){
        cin>>n>>v;
        for(int i=1;i<=n;i++)
            cin>>Bag[i].c;
        for(int i=1;i<=n;i++)
            cin>>Bag[i].v;
        memset(f,0,sizeof(f));
        for(int i=1;i<=n;i++)
            for(int j=v;j>=Bag[i].v;j--)
            f[j]=max(f[j],f[j-Bag[i].v]+Bag[i].c);
        cout<<f[v]<<endl;
    }
    return 0;
}