HDU 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43332    Accepted Submission(s): 19229

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														
标准的DFS，深度搜索一下就好，我试了，要是输入个18，得跑好久，估计，是跑不出了，下面的代码可以ac
#include <stdio.h>
#include <string.h>

int pri[55],cnt=1;
int a[25],jdg[25],n;
void dfs(int tmp)
{
	int i;
	
	if(tmp>n)
	{
		if(pri[a[n]+a[1]]&&pri[a[n]+a[n-1]])
		{
			
			for(i=1;i<=n;i++)
			{
				if(i==n)
					printf("%d\n",a[i]);
				else
					printf("%d ",a[i]);
			}
		}
		
		return ;
	}
	if(tmp>=3&&(!pri[a[tmp-1]+a[tmp-2]]))
	{
		return ;
	}
	for(i=2;i<=n;i++)
	{
		if(!jdg[i])
		{
			a[tmp]=i;
			jdg[i]=1;
			dfs(tmp+1);
			jdg[i]=0;
		}
	}	
}
int main()
{
	int i,j;
	
	for(i=0;i<55;i++)
		pri[i]=1;
	for(i=2;i<=30;i++)
	{
		if(pri[i])//pri[i]=0
		{
			for(j=i+i;j<55;j=j+i)
				pri[j]=0;
		}
	}
	pri[0]=pri[1]=0;
	
	while(~scanf("%d",&n))
	{
		memset(jdg,0,sizeof(jdg));
		a[1]=1;
		jdg[1]=1;
		printf("Case %d:\n",cnt++);
		dfs(2);
		printf("\n");
		
	}
	
	return 0;
}