【dfs】hdu 1016 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

图片就不放了。。做下笔记以后回来复习。。
code：

#include <iostream>#include <cmath>#include <cstring>using namespace std;int n,a[25],book[25];int prime(int x){    int i;    for(i=2;i<=sqrt(x);i++)        if(x%i==0) break;    if(i<=sqrt(x)) return 0;    else return 1;}void dfs(int step){    int first=1;    a[n+1]=a[1];    a[1]=1;    book[1]=1;    if(step==n+1) {        if(prime(a[n]+a[1])==0) return ;//判断最后一个数和第一个数的和是不是素数        for (int i=1;i<=n;i++){            if(first) first=0;            else cout<<' ';            cout<<a[i];        }        cout<<endl;    }   //用递归实现dfs，满足条件输出，不满足条件直接返回        for(int i=2;i<=n;i++){            if(book[i]==0 && prime(i+a[step-1])){  //在循环的时候判断，避免超时，减少运算次数。。                a[step]=i; //第一个位置放的数为i                book[i]=1; //表示已经用过i这个数                dfs(step+1);//递归调用                book[i]=0; //用完后还原，下次搜索时仍能访问            }        }    return ;}int main(){    int sum=1;    memset(a,0,sizeof(a));    while(cin>>n){            cout<<"Case "<<sum<<':'<<endl;        sum++;        if(n%2==1) cout<<endl; //剪枝，不然会超时，数据太大        else {            memset(book,0,sizeof(book));            dfs(2);            cout<<endl;        }    }    return 0;}