hdu 1016 Prime Ring Problem

hdu 1016 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：简单来说，如果用暴力把每一种结果都测试一遍，那么排列的总数会很大，明显会超时。因此，需要用到回溯法，详细见以下代码：

#include<iostream>using namespace std;int visit[20],a[21],n;int _prime(int p)           //判断素数{    int i;    for(i=2;i*i<=p;i++)      if(p%i==0)        return 0;    return 1;}void dfs(int k,int num,int n)    //递归广度优先搜索{    a[k]=num;    visit[num]=1;    if(k==n&&_prime(a[1]+a[k]))      //边界条件    {        cout<<"1";        for(int i=2;i<=n;i++)          cout<<" "<<a[i];        cout<<endl;    }    else for(int i=1;i<=n;i++)    {        if(!visit[i]&&_prime(i+a[k]))        {            dfs(k+1,i,n);            visit[i]=0;        //清除标志        }     }}int main(){    int t=1;    while(cin>>n)    {        a[1]=1;        cout<<"Case "<<t<<":"<<endl;        memset(visit,0,sizeof(visit));        dfs(1,1,n);        cout<<endl;        t++;    }    return 0;}