背包问题—杭电2602 Bone Collector

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): 25797    Accepted Submission(s): 10451

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
   
   
   
   

 

#include <iostream>
using namespace std;
#define N 1001
int dp[N][N];
int c[N],w[N];
int max(int a,int b)
{
	return a>b?a:b;
}
int main()
{
	int cishu,k;
	cin>>cishu;
	for(k=1;k<=cishu;k++)
	{
		int n,v;
		cin>>n>>v;
		int i,j;
		for(i=1;i<=n;i++)
			cin>>c[i];
		for(i=1;i<=n;i++)
			cin>>w[i];
		memset(dp,0,sizeof(int));
		for(i=1;i<=n;i++)
			for(j=0;j<=v;j++)
			{
				if(w[i]<=j)
                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+c[i]);
				else
					dp[i][j]=dp[i-1][j];					
			}
			cout<<dp[n][v]<<endl;
	}
	return 0;
}