抛抛DFS——— Prime Ring 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

 

 

这题运用了DFS，但是又与前面的有些不同，首先我用了两个全局的数组，一个存值，一个是标记数组。

在后面DFS函数中，递归的时候，我运用了回溯。吴涛说那里用回溯，我一直没有想到怎么用，但是后来的后来，看了一些东西，有了感觉，理解了很多。每次递归，等递归的最后结束语句执行以后，都会进行回溯，将前面的值回溯出来，递归进去是从前往后，但是递归结束，进行回溯的时候是从后往前的。

 

 

#include<stdio.h>
#include<stdlib.h>
#include<math.h> 
int pao[30], flag[25], n;

int prime( int x )
{
    for( int i = 2; i <= sqrt( x ); i++ )
      if( x % i == 0 )
         return 0;
      return 1;
}

void DFS( int x )
{
     int i, j;
     if( ( x == n ) && prime( pao[x] + pao[1] ) && prime( pao[x] + pao[x-1] ) )
     {
          for( i = 1; i <= n; i++ )
            printf( i==1 ? "%d" :" %d", pao[i] );
          puts("");
     }
    else 
      {
            for( i = 2; i <= n; i++)
               if( prime( pao[x]+i ) && !flag[i] )
               {
                     flag[i] = 1;//标记 
                     pao[x+1] = i;
                     DFS( x+1 );
                     flag[i] = 0;//回溯 
               }
       }
}
int main()
{
    int k = 1;
    while( scanf( "%d", &n ) == 1 )
    {
           printf( "Case %d:\n", k++ );
           for( int i = 0; i < n; i++ )
             {
                      pao[i] = 0;
                      flag[i] = 0;
             }
             pao[1] = 1;
             DFS( 1 );
             puts( "" ); 
    }
    return 0;
}