杭電1016-Prime Ring Problem

Prime Ring Problem

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

Total Submission(s): 19646    Accepted Submission(s): 8793

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
  
  
   經典DFS題，這裡注意輸出格式就能AC了：
  
  
   
          
#include<iostream>
#include<cmath>
const int MAX=100;
int prim[MAX];
int chucun[MAX];
int mark[MAX];
int n;
int count;
using namespace std;
int main()
{
	int i,p=1;
	void DFS(int x);
	bool prime(double x);
	while(cin>>n)
	{
		cout<<"Case "<<p++<<':'<<endl;
		chucun[0]=1;
		memset(mark,0,sizeof(mark));
		DFS(1);
		cout<<endl;
	}
}
bool prime(double x)
{
	int i;
	for(i=2;i<=sqrt(x);i++)
	{
		if(int(x)%i==0)
		return false;
	}
	return true;
}
void DFS(int x)
{
		if(x==n&&(prime(double(1+chucun[x-1]))))
		{
			for(int i=0;i<x;i++)
			{
				if(i==0)
				cout<<chucun[i];
				else
				cout<<" "<<chucun[i];
			}
			cout<<endl;
		}
		else
		{
			for(int i=2;i<=n;i++)
			{
				if(mark[i]||!prime(double(chucun[x-1]+i)))
				continue;
				mark[i]=1;
				chucun[x]=i;
				DFS(x+1);
				mark[i]=0;
			}
		}
}