hdu 1016 Prime Ring Problem（素数环）

Prime Ring Problem

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

解析：本题在lnj的紫书上面作为例题讲过，不过主要体现的是回溯，在这里，我重点就广搜这方面来讲。

思路：首先需要开一个标记数组和一个判断素数的调用函数，然后从第二个位置开始进行广搜，从2到n，如果找到符合条件的数，则用标记数组进行标记，一直递归搜索，知道第n个位置，如果符合条件跳出边界输出结果，如果中途不符合条件则清除当前节点标记并一层一层返回上一层，不断回溯。

具体解释看如下代码：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int n,a[22],vis[22];       //a为素数数组，vis为标记数组 int prime(int a)          //判断素数 {    for(int i=2;i*i<=a;i++)      if(a%i==0)        return 0;    return 1;}void dfs(int x){     if(x==n+1&&prime(a[x-1]+a[1]))     //判断边界条件，如果满足即输出     {        for(int i=1;i<n;i++)          cout<<a[i]<<" ";        cout<<a[n]<<endl;    }    else    {        for(int i=2;i<=n;i++)        //遍历寻找符合的数         {            if(!vis[i]&&prime(i+a[x-1]))             {                vis[i]=1;            //对当前满足的数所对应的标记下标数组进行标记                 a[x]=i;              //对当前满足的数所对应的素数下标数组进行赋值                 dfs(x+1);           //递归广度优先搜索                 vis[i]=0;           //回溯时注意清除标记             }        }    }}int main(){    int t=0;    while(cin>>n)    {        t++;        cout<<"Case "<<t<<":"<<endl;        memset(vis,0,sizeof(vis));        a[1]=1;        dfs(2);      //从第二个位置开始递归搜索         cout<<endl;    }    return 0;}