背包问题

杭电ACM 2014暑期集训队——选拔安排~

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25800    Accepted Submission(s): 10453

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

 

#include<iostream>
using namespace std;
int max(int a,int b)
{if(a>b)
return a;
else return b;}
int n,j,v,c[1004],w[1004],dp[1004][1004],k,i;
int main()
{
	
	cin>>k;
	while(k--)
	{
		memset(dp,0,sizeof(dp));
		cin>>n>>v;
		for(i=1;i<=n;i++)
			cin>>c[i];
		for(i=1;i<=n;i++)
			cin>>w[i];
		
		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;
}