hduProblem-1016简单dfs

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 49746    Accepted Submission(s): 21932

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
  
  

AC源码：

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

const int MAXN=25;
int flag[MAXN];
bool is_prime(int x)
{
	for(int i=2;i*i<=x;++i)
		if(x%i==0)
			return false;
	return true;
}
void dfs(int cur,int n,vector<int> &A)
{
	if(cur==n)
	{
		if(is_prime(A[0]+A[cur-1]))
		{
			int flg=1;
			for(int i=0;i<n;++i)
			{
				if(flg)
				{
					cout<<A[i];flg=0;
				}
				else
					cout<<" "<<A[i];
			}
			cout<<endl;
		}
		return ;
	}
	for(int i=2;i<=n;++i)
	{
		if(flag[i]&&is_prime(A[cur-1]+i))
		{
			A.push_back(i);
			flag[i]=0;
			dfs(cur+1,n,A);
			A.pop_back();	//恢复原样
			flag[i]=1;	//恢复原样
		}
	}
}

int main()
{
	int n,kase=0;
	while(cin>>n)
	{
		cout<<"Case "<<(++kase)<<":"<<endl;
		for(int i=1;i<=n;++i)
			flag[i]=1;
		vector<int> A;
		A.push_back(1);
		dfs(1,n,A);
		cout<<endl;
	}
	return 0;
}