素数环（C - 暴力求解）

素数环（暴力）（紫书194页）

Description

 

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 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 <= 16))
-->

n (0 < n <= 16)

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.

You are to write a program that completes above process.

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

题意：
输入正整数n，把1—n组成一个环，使相邻的两个整数为素数。输出时从整数1开始，逆时针排列。恰好能构成一个环输出一次，n (0 < n <= 16)

分析：
每个环对应于1——n的一个排列，但排列总数高达16！=2*10^13，普通写法会超时，应用回溯法。
DFS。深度优先遍历。

代码：

 #include<cstdio>
 #include<iostream>
 using namespace std;

 int n, A[], isp[], vis[];

 int is(int x)
 {
     for(int i = ; i*i <= x; i++)
         if(x % i == )
             return ;
         return ;
 }

 void dfs(int cur)
 {
     if(cur==n&&isp[A[]+A[n-]])
     {
         for(int i=;i<n;i++)
             printf("%d ",A[i]);
         printf("\n");
     }
     else for(int i=;i<=n;i++)
         if(!vis[i]&&isp[i+A[cur-]])
         {
             A[cur] = i;
             vis[i] = ;
             dfs(cur+);
             vis[i] = ;
         }
 }

 int main()
 {
     //int n;
     int m=;
     while(scanf("%d",&n)!=EOF)
     {
     for(int i=;i<=n*;i++)
         isp[i] = is(i);
     memset(vis, , sizeof(vis));
     A[] = ;
     m++;
     printf("Case %d:\n",m);
     dfs();
     printf("\n");
     }
     return ;
 }