素数环问题，深度搜索

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.<br><br>Note: the number of first circle should always be 1.<br><br><img src=../../data/images/1016-1.gif><br>

 

Input

n (0 < n < 20).<br>

 

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.<br><br>You are to write a program that completes above process.<br><br>Print a blank line after each case.<br>

 

Sample Input

   
   
   
   

    
    
    
    6<br>8<br>
   
   
   
   
   
   
   
   

    
    
    
    
 
     
      
      
      
      

       
       
       
       Case 1:<br>1 4 3 2 5 6<br>1 6 5 2 3 4<br><br>Case 2:<br>1 2 3 8 5 6 7 4<br>1 2 5 8 3 4 7 6<br>1 4 7 6 5 8 3 2<br>1 6 7 4 3 8 5 2<br>
      
      
      
      
      
      
      
      


      
      
      
      
 
    
题意：给出一个正整数，第一个数一定是一，然后搜索怎么排列能做成素数的环。
   
   
   
   
   
   
   
   

    
    
    
    思路：就是广度搜索，符合要求就输出这一序列。
   
   
   
   
   
   
   
   

    
    
    
    #include<iostream>
    
    
    
    
#include<cstring>
    
    
    
    
using namespace std;
    
    
    
    
int n,cnt = 1;
    
    
    
    
int vis[21],A[21];//存放进这之中
    
    
    
    
int primelist[38] = {0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1};//因为0<n<20所以这一范围内，素数就那么几个，这样搜索比较快
    
    
    
    
void dfs(int cur)//深度搜索
    
    
    
    
{
    
    
    
    

    
    
    
     if(cur == n && primelist[A[0]+A[n-1]])//判断是否是素数
    
    
    
    

    
    
    
     {
    
    
    
    

    
    
    
     cout << A[0] ;
    
    
    
    

    
    
    
     for(int i = 1;i < n;++i)
    
    
    
    

    
    
    
     cout << " "<< A[i];
    
    
    
    

    
    
    
     cout <<endl;
    
    
    
    

    
    
    
     }
    
    
    
    

    
    
    
     else for(int i = 2;i <= n;++i)
    
    
    
    

    
    
    
     if(!vis[i] && primelist[i+A[cur-1]])
    
    
    
    

    
    
    
     {
    
    
    
    

    
    
    
     A[cur] = i;
    
    
    
    

    
    
    
     vis[i] = 1;
    
    
    
    

    
    
    
     dfs(cur + 1);
    
    
    
    

    
    
    
     vis[i] = 0;
    
    
    
    

    
    
    
     }
    
    
    
    
}
    
    
    
    

    
    
    
    

    
    
    
    
int main()
    
    
    
    
{
    
    
    
    

    
    
    
     A[0] = 1;
    
    
    
    

    
    
    
     while(cin >> n)
    
    
    
    

    
    
    
     {
    
    
    
    

    
    
    
     memset(vis,0,sizeof(vis));
    
    
    
    

    
    
    
     cout << "Case " << cnt++ <<":\n";
    
    
    
    

    
    
    
     dfs(1);
    
    
    
    

    
    
    
     cout << endl;
    
    
    
    

    
    
    
     }
    
    
    
    

    
    
    
    

    
    
    
    

    
    
    
     return 0;
    
    
    
    
}