HDOJ 1016 Prime Ring Problem(dfs)

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 34066 Accepted Submission(s): 15082



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

题意：由1到n的n个数，连接成一个圆环，每两个相邻的数相加必须是素数，输出这些数的序列。



代码如下：

#include<cstdio>
#include<cstring>
int mark[22],a[22];
int n;

int prime(int x)//判断素数 
{
	int i;
	if(x==1)
	  return 0;
	if(x==2||x==3)
	  return 1;
	for(i=2;i*i<=x;i++)
	{
		if(x%i==0)
		  return 0;
	}
	return 1;
}

void dfs(int num,int n)
{
	int i;
	if(num==n+1&&prime(a
+1))
	{
		for(i=1;i<n;i++)
		  printf("%d ",a[i]);
		printf("%d\n",a
);
	}
	else
	{
		for(i=2;i<=n;i++)
		{
			if(!mark[i]&&prime(a[num-1]+i))
			{
				a[num]=i;
				mark[i]=1;//标记已经找过的数 
				dfs(num+1,n);
				mark[i]=0;//查找结束，或失败时，回溯还原上一个被标记的点 
			}
		}
	}
}

int main()
{
	int i,t=0;
	while(scanf("%d",&n)!=EOF)
	{
		memset(a,0,sizeof(a));
		memset(mark,0,sizeof(mark));
		a[1]=1;mark[1]=1;
		printf("Case %d:\n",++t);
	    if(n==1)
	        printf("1\n");
	    else
	    	dfs(2,n);
	    printf("\n");
	}
	return 0;
}