HDU1016:Prime Ring Problem(DFS)

Problem Description

A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

Note: the number of first circle should always be 1.

 

Input

n (0 < n < 20).

 

Output

The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.

You are to write a program that completes above process.

Print a blank line after each case.

 

Sample Input

   
   
   
   

    
    
    
    6
8
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case 1:
1 4 3 2 5 6
1 6 5 2 3 4

Case 2:
1 2 3 8 5 6 7 4
1 2 5 8 3 4 7 6
1 4 7 6 5 8 3 2
1 6 7 4 3 8 5 2
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    素数筛选法+深搜
上代码：
    
    
    
    
#include<iostream>
#include<cstring>

using namespace std;

int cnt;
int a[22];
int book[22];
int prime[50];
int n;
int Case;
void isPrime()
{
	for(int i=1;i<50;i++)
	prime[i]=1;
	
	prime[0]=prime[1]=0;
	
	for(int i=2;i<50;i++)
	{
		if(prime[i])
		{
			for(int j=2*i;j<50;j+=i)
			prime[j]=0;
		}
		
	}
	
	 
}

void dfs(int step)
{
	if(step==n+1&&prime[a[n]+1])
	{
		
		cnt++;
	//	cout<<"Case "<<Case<<":"<<endl;
		for(int i=1;i<n;i++)
		{
			cout<<a[i]<<" ";
		}
		cout<<a[n]<<endl;
		
		
	}
	
	for(int i=2;i<=n;i++)
	{
		if(!book[i]&&prime[i+a[step-1]])
		{
			a[step]=i;
			book[i]=1;
			dfs(step+1);
			book[i]=0;
			
		}
	}
	return ;
}
int main()
{
	isPrime();
	while(cin>>n)
	{
		memset(book,0,sizeof(book));
		Case++;
		cout<<"Case "<<Case<<":"<<endl;
		a[1]=1;
		dfs(2);
		cout<<endl;
	}
	
	return 0;
}