hdu 1016 Prime Ring Problem（深度优先搜索）

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 12105    Accepted Submission(s): 5497

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

题意：输入一个 n 找出1~n的组合，使得相邻两个数之和为素数；

分析：预处理40之间的素数，然后回溯；

 #include<iostream>
 #include<cstring>
 #define N 25
 #define M 40
 using namespace std;

 bool is_prime[M],visited[N];
 int n,test,ans[N];

 void work(int k)
 {
     int i;
     if(k==n+)
     {
         if(!is_prime[ans[n]+ans[]]) return ;
         for(i=;i<=n-;i++)
             cout<<ans[i]<<" ";
         cout<<ans[i]<<endl;
         return ;
     }
     for(i=;i<=n;i++)
     {
         if(!visited[i]&&is_prime[ans[k-]+i])
         {
             visited[i]=true;
             ans[k]=i;
             work(k+);
             visited[i]=false;
         }
     }
 }

 bool prime(int n)
 {
     if(n==) return false;
     if(n==||n==) return true;
     int i;
     for(i=;i<n;i++)
         if(n%i==)
             return false;
     return true;
 }

 int main()
 {
     int i;test=;
     for(i=;i<M;i++) is_prime[i]=prime(i);
     while(cin>>n)
     {
         ans[]=;
         memset(visited,false,sizeof(visited));
         cout<<"Case "<<test<<":"<<endl;
         work();
         test++;
         cout<<endl;
     }
     return ;
 }