hdoj-1016 Prime Ring Problem

Prime Ring Problem

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

Total Submission(s): 54412    Accepted Submission(s): 24090

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
  
  

  题目链接

题意：输入一个n，将这n个数进行排列，使得相邻两个数之和为素数。

#include<iostream>
#include<cstdio>
using namespace std;
int num[21],mark[21];
int n;
int p[12]={2,3,5,7,11,13,17,19,23,29,31,37};
int prime(int a)
{
	int i;
	for(i=0;i<12;i++)
	{
		if(a==p[i])
			return 1;
	}
	return 0;
}
void print_num()
{
	int i;
	for(i=1;i<n;i++)
		printf("%d ",num[i]);
	printf("%d",num[n]);
}
int dfs(int pre,int post,int flag)
{
	int i;
	if(!prime(pre+post))
	{
		return 0;
	}
	num[flag]=post;
	if(flag==n&&prime(post+1))
	{
		print_num();
		printf("\n");
		return 1;
	}
	mark[post]=0;
	for(i=2;i<=n;i++)
		if(mark[i]!=0&&dfs(post,i,flag+1))
			break;
	mark[post]=1;
	return 0;				
}
int main()
{
	int count,i;
	count=1;
	while(scanf("%d",&n)!=EOF)
	{
		for(i=1;i<=n;i++)
		{
			mark[i]=i;
		}
		num[1]=1;
		printf("Case %d:\n",count++);
		if(n==1)
			printf("1\n");
		for(i=2;i<=n;i++)
			dfs(1,i,2);
		printf("\n");
	}
	return 0;
}