poj1016 Prime Ring Problem---dfs

Prime Ring Problem

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

Total Submission(s): 22718 Accepted Submission(s): 10116



[align=left]Problem Description[/align]
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.



[align=left]Input[/align]
n (0 < n < 20).

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

6
8

[align=left]Sample Output[/align]

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

[align=left]Source[/align]
Asia 1996, Shanghai (Mainland China)

题目大意：给你一个n，输出用1~n的数组成一个环，相邻两数之间的和为素数。

代码：

#include <iostream>
#include <cstring>
using namespace std;

int hash[30];
int arr[100];
int n;

int check(int x)
{
for(int i=2;i<x;i++)
{
if(x%i==0)
return 0;
}
return 1;
}

void dfs(int order,int index)
{
int i;
arr[order]=index;
hash[index]=1;

if(order==n)
{
if(check(arr[order]+arr[1]))
{
cout<<"1";
for(i=2;i<=n;i++)
cout<<" "<<arr[i];
cout<<endl;
}
return;
}

for(i=1;i<=n;i++)
{
if(!hash[i]&&check(arr[order]+i))
{
dfs(order+1,i);
hash[i]=0;
}
}
return;
}
int main()
{
int flag=1;
while(cin>>n)
{
cout<<"Case "<<flag++<<":"<<endl;

memset(hash,0,sizeof(hash));
dfs(1,1);

cout<<endl;
}
return 0;
}