HDU 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52000    Accepted Submission(s): 23028

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

 

一道经典的深搜题，题意应该很容易看懂吧.........

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int n,num[21],book[21];//最大输入不超过20，也就是说能达到的最大的素数也就是19+18=37int prime[12]={2,3,5,7,11,13,17,19,23,29,31,37};bool isprime(int x){for(int i=0;i<12;i++)if(x==prime[i])return true;return false;}void dfs(int step){//记得要判断第一个点和最后一个点的和是不是素数if(step==n&&isprime(1+num[step])){for(int i=1;i<=n-1;i++) printf("%d ",num[i]);printf("%d\n",num[n]);}else{for(int i=2;i<=n;i++){if(!book[i]&&isprime(i+num[step])){num[step+1]=i;book[i]=1;dfs(step+1);book[i]=0;}}}}int main(){int count=1;while(~scanf("%d",&n)){num[1]=1;memset(book,0,sizeof(book));printf("Case %d:\n",count++);dfs(1);cout<<endl;}return 0;}