HDU1016 Prime Ring Problem (经典的深搜)

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

/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

题意为两个相邻的圆圈的数值之和为素数

#include<iostream>
using namespace std;
int P[38]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1};
int visited[21];
int a[21];
int n;
void DFS(int count)
{
	int i;
	if(count==n&&P[a[count-1]+1]==1)
	{
		cout<<a[0];
		for( i=1;i<n;i++)
		{
			cout<<' '<<a[i];
		}
		cout<<endl;
	}
	else
	{
		for(i=2;i<=n;i++)
		{
			if(!visited[i]&&P[a[count-1]+i]==1)
			{
				a[count]=i;
				visited[i]=1;
				DFS(count+1);
				visited[i]=0;
			}
		}
	}
}
int main()
{
	a[0]=1;
	int A=1;
	while(cin>>n)
	{
		cout<<"Case "<<A++<<':'<<endl;
		DFS(1);
		cout<<endl;
	}
	return 0;
}