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;
}