杭电1016————素数环之DFS

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14061    Accepted Submission(s): 6390

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

这个题目很容易想到搜索的办法，是属于比较简单的DFS题目，代码经过测试AC，但是时间用了大约500MS，亲测当n>=11时，这段代码卡顿。杭电的范围是0<n<20，目测是数据水了。杭电最快是15MS。
#include <stdio.h>
#include <string.h>
#include <math.h>
#define maxn 30
int prime[maxn],vis[maxn];/*prime记录要输出的序列 vis表示当前数字是否已被使用*/ 
int n;/*the number*/
void print()
{
	for(int i = 0;i < n;i++)
	{
		if(i != n-1)
			printf("%d ",prime[i]);
		else
			printf("%d\n",prime[i]);
	}
}
int isprime(int x)
{
	int flag = 1;
	for(int i = 2;i <= sqrt(x);i++)
	{
		if(x % i == 0)
		{
			flag = 0;
			break;
		}
	}
	return flag;
}
void DFS(int x,int count)/*x记录数字的值,count记录第几个数*/ 
{
	if( isprime(x+1) && count == n)/*终止条件,即递归出口*/ 
	{
		print();
		return ;
	}
	else
	{
		for(int i = 2 ; i <= n; i++)
		{
			if( isprime(i+x) && vis[i] == 0)
			{
				vis[i] = 1;
				prime[count++] = i;
				DFS(i,count);
				count--;
				vis[i] = 0;
			}
		}	
	}
}
int main()
{
	
	int cases = 0;/*case x*/
	
	while(scanf("%d",&n) != EOF)
	{
		cases++;
		memset(prime,0,sizeof(prime));
		memset(vis,0,sizeof(vis));
		prime[0] = 1;/*the begining number is always one*/
		vis[1] = 1;/*one has been used*/ 
		printf("Case %d:\n",cases);
		DFS(1,1);
		printf("\n");
	} 
	return 0;
}