Prime Ring Problem HDU - 1016 (dfs)

Prime Ring Problem

HDU - 1016

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.

 

Inputn (0 < n < 20). 
OutputThe 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

 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<queue>
 
 using namespace std;
 
 int n;
 int a[];
 int vis[];
 bool mark[];
 
 void init()        // 素数筛
 {
     for(int i = ; i < ; ++i)
         mark[i] = true;
 
     for(int i = ; i < ; ++i)
     {
         if(mark[i] == true)
         {
             for(int j = i*i; j < ; j += i)
             {
                 mark[j] = false;
             }
         }
     }
 }
 
 // 边枚举边判断，不要最后一次性判断，会超时
 void dfs(int step)
 {
     if(step > )
     {
         if(mark[a[step-]+a[step-]] == false)    // 判断最后两个数
             return;
     }
 
     if(step == n+)
     {
         if(mark[a[n]+a[]] == false)    // 判断最后一个数与第一个数
             return;
         for(int i = ; i < n; ++i)
             printf("%d ", a[i]);
         printf("%d\n", a[n]);
 
     }
 
     for(int i = ; i <= n; ++i)
     {
         if(!vis[i])
         {
             a[step] = i;
             vis[i] = ;
             dfs(step+);
             vis[i] = ;
         }
 
     }
 }
 
 int main()
 {
      init();
      int cas = ;
      a[] = ;
      while(scanf("%d", &n) != EOF)
      {
          printf("Case %d:\n", cas++);
          memset(vis, , sizeof(vis));
          dfs();
          printf("\n");
      }
 
     return ;
 }