Q - 17 深度搜索

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：

给你一个整数n，从1到n组成一个圆环，要求相邻的两个数的和为素数，让你输出每种情况！！

思路：

运用深度搜索，对每个位置的数值进行判断，可以定义一个判断是否素数的函数（因个数较为小，可以直接写出）和输出的函数，即当找的数的位数与n相等时以及最后一位和第一位和为素数，输出！！还有就是每次情况结束之后，要进行回溯！来两个数组，一个用来存放每个位置存放的数，另一个记录每个数的状态！！！

代码：

#include <bits/stdc++.h>
using namespace std;
int     prime[12]={2,3,5,7,11,13,17,19,23,29,31,37};
int n,pos[21],con[21];
bool isPrime(int a,int b)
{
	int c=a+b;
	for(int i=0;i<12;i++)
	if(prime[i]==c)
	return 1;
	return 0;
}
void output()
{
	for(int i=1;i<n;i++)
	cout<<pos[i]<<' ';
	cout<<pos[n]<<endl;
}
void go(int m)
{
	for(int i=2;i<=n;i++)
	{
		if(isPrime(pos[m-1],i)&&con[i]==0)
		{
			pos[m]=i;
			con[i]=1;	if(m==n&&isPrime(1,pos[m]))	
			{
				output();
			}
	else  go(m+1);
	con[i]=0;
		}
	}
}
int main()
{
	int ans=0;
	pos[1]=1;
	while(cin>>n)
	{	memset(con,0,sizeof(con));
		con[1]=1;
		ans++;
		cout<<"Case "<<ans<<":"<<endl;
		go(2);
		cout<<endl;
	}
	return 0;
}

搜索题还是吃力！！多刷题！！！！