01-背包问题---Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 35547    Accepted Submission(s): 14631

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
   
   
   
   

 
未进行优化的代码：

#include <iostream>
#include <cstring>
using namespace std;
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
int max(int a,int b)
{
	return (a>b)?a:b;
}
int val[1005],vol[1005];
int v[1005][1005];
int main(int argc, char *argv[]) {
	int t;
	int n,vv;
	
	cin>>t;
	while(t--)
	{
		cin>>n>>vv;
		for(int i=1;i<=n;i++)
		cin>>val[i];
		for(int i=1;i<=n;i++)
		cin>>vol[i];
		for(int i=0;i<=n;i++)
		v[i][0]=0;
		for(int j=0;j<=vv;j++)
		v[0][j]=0;
		memset(v,0,sizeof(v));
		int i,j;
		for(i=1;i<=n;i++)
		{
			for(j=0;j<=vv;j++)
			{
				if(vol[i]<=j)
				{
					v[i][j]=max(v[i-1][j],v[i-1][j-vol[i]]+val[i]);
				}
				else
				v[i][j]=v[i-1][j];
			}
		}
		cout<<v[n][vv]<<endl;
	}
	return 0;
}

优化后：

#include <iostream>
#include <cstring>
using namespace std;
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
int max(int a,int b)
{
	return (a>b)?a:b;
}
int val[1005],vol[1005];
int v[1005];
int main(int argc, char *argv[]) {
	int t;
	int n,vv;
	
	cin>>t;
	while(t--)
	{
		cin>>n>>vv;
		for(int i=1;i<=n;i++)
		cin>>val[i];
		for(int i=1;i<=n;i++)
		cin>>vol[i];
		for(int i=0;i<=vv;i++)
		v[i]=0;
		
		memset(v,0,sizeof(v));
		int i,j;
		for(i=1;i<=n;i++)
		{
			for(j=vv;j>=0;j--)
			{
				if(vol[i]<=j)
				{
					v[j]=max(v[j],v[j-vol[i]]+val[i]);
				}
			}
		}
		cout<<v[vv]<<endl;
	}
	return 0;
}