NOJ——[1361] Prime Again

• 问题描述
• We all vary familar with prime numbers, now we get a integer N（1<=n <= 16）, for 1 to N, your task is to write a 
  Cyclic sequences a1~an, all numbers only use once, and every sum of adjacent two numbers should be prime. No matter how you cut a circle, they're just the same, so one circle should be print only once. For all results, we print by lexicographical order. If there's no answer, you needn't print anything.
• 输入
• There are many cases, end until EOF, each a integer N（1<=n <= 16） in a single line.
• 输出
• Please prnit by lexicographical order.
• 样例输入

• 4
  6
  

• 样例输出

• 1 2 3 4
  1 4 3 2
  1 4 3 2 5 6
  1 6 5 2 3 4
  

• 提示

• 无

• 来源

• Minary

素数环，dfs回溯解决，特判n==1。

#include<stdio.h>
#include<string.h>

bool prime[100];

int ans[20],n;

bool vis[100];

void dfs(int cur)
{
	if(cur == n && prime[ans[0]+ans[n-1]])
	{
		for(int i=0;i<n;i++)
        {
        	printf("%d",ans[i]);
        	if(i<n-1)
        	  printf(" ");
        }
        printf("\n");
        return ;
	}
	for(int i=2;i<=n;i++)
	{
		if(!vis[i] && prime[ans[cur-1]+i])
		{
			vis[i]=1;
			ans[cur]=i;
			dfs(cur+1);
			vis[i]=0;
		}
	}
}

int main()
{
	memset(prime,0,sizeof(prime));
	prime[2]=1;
	prime[3]=1;
	prime[5]=1;
	prime[7]=1;
	prime[11]=1;
	prime[13]=1;
	prime[17]=1;
	prime[19]=1;
	prime[23]=1;
	prime[29]=1;
	prime[31]=1;
	while(~scanf("%d",&n))
	{
		memset(vis,0,sizeof(vis));
		if(n>1)
		  for(int i=1;i<=n;i++)
		  {
			  vis[i]=1;
			  ans[0]=i;
			  dfs(1);
		  }
		
	}
	return 0;
}