Prime Ring Problem 1016（抽象DFS）

题目链接：Prime Ring Problem

题目：

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

题意：
就是让你输入一个数n，然后从1-n这n个数，依次逆时针摆放，形成一个圆（如上图所示）。要求，任意相邻的两个数相加起来的和一定为素数；注意：第一个数从1开始；

AC代码：

#include 

using namespace std;
int n,ans=1;
int p[22];//将合适的结果记录下来
bool vist[22];//标记数组；
bool check(int a)  //检查相邻两个数加起来是否为素数
{
    if(a<=1) return false;
    for(int i=2;i*i<=a;i++)
    {
        if(a%i==0)
            return false;
    }
    return true;
}
void dfs(int x)
{
    if(x==n&&check(1+p[n]))//当最后一个圈中放第n个数字时，并且这第n个数字能满足和第一个数字相加为素数就是符合条件
    {
        for(int i=1;i<=n;i++)
        {
            if(i==1) cout <<p[i];
            else {
                cout <<" "<<p[i];
            }
        }
        cout <<endl;
        return ;
    }
    for(int i=2;i<=n;i++)
    {
        if(check(i+p[x])&&!vist[i])//检测从2-n的每一个数是否符合条件
        {
            p[x+1]=i;//如果符合条件，则让存放数组的下一个地方存放该数字
            vist[i]=true;//把该数字进行标记
            dfs(x+1);//进行下一次的dfs
            vist[i]=false;//恢复标记，回溯
        }
    }
}
int main()
{
    while(cin >>n)
    {
        memset(p,0,sizeof(p));
        memset(vist,false,sizeof(vist));
        p[1]=1; //第一个数一定是1，所以直接存放
        cout <<"Case "<<ans++<<":"<<endl;
        dfs(1);//从1开始进行dfs
        cout <<endl;
    }

    return 0;
}