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

68

 

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 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; }