hdu 1016 Prime Ring Problem

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
  

 

Source

Asia 1996, Shanghai (Mainland China)

 

BFS搜索

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
bool visit[22];//访问标记数组
int n;//代表总数
int map[22];//存放行动路径
bool isprime[41]={0,0,0,1,0,1,0,1,0,0,0,
                    1,0,1,0,0,0,1,0,1,0,
                    0,0,1,0,0,0,0,0,1,0,
                    1,0,0,0,0,0,1,0,0,0
                 };//前40项素数状况
void dfs(int pos) 
{
	int i;
	if(pos==n)//搜索到底了
	{
		if(isprime[map[n-1]+1])//判断最后一个和第一个能否构成环
		{
			 for(i=0;i<n-1;i++)
			 {
			 	 printf("%d ",map[i]);
			 } 
			printf("%d\n",map[n-1]);
		} 
		return ;//不满足情况就返回结束 
	} 
	//没有搜索到底时，就接着搜索
	for(i=1;i<=n;i++)
	{
		 if(!visit[i]) //已经被访问
		   continue;
		 if(!isprime[i+map[pos-1]])//不是素数相邻 
		  continue;
		 visit[i]=false;//标记被访问
		 map[pos]=i;//加入路径
		 dfs(pos+1);//接着搜索
		 
		 //递归结束之后 
		 visit[i]=true; //改变的要变回来 
		   
	} 
	 
}
int main()
{
	int k;
	k=0;
	while(scanf("%d",&n)!=EOF)
	{
		memset(visit,true,sizeof(visit));
		map[0]=1;//都是从一开始的
		visit[1]=false;//标记已经访问
		printf("Case %d:\n",++k);
		dfs(1);//从一开始搜索 
		printf("\n");
	}
return 0; 
}