Prime Ring Problem(dfs水)

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1016

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34554    Accepted Submission(s): 15303

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)

 简单的dfs

直接上代码:

这里介绍一个报错:Floating point exception (core dumped) linux下报这个一般就是出现了除0或者模0操作....写代码要仔细呀

 #include<cstdio>
 #include<cstring>
 using namespace std;
 #define N 50
 int a[N];
 bool vis[N];
 bool is_prime[N];
 void init()
 {
     for(int i = ;i < N ;i++) is_prime[i] = ;
     for(int i =  ;i < N ;i++)
     {
         for(int j =  ; j <= i/ ;j++)
         {
             if(i%j==) {is_prime[i] =  ; continue;}
         }
     }
 }
 void dfs(int n, int cnt)
 {
     if(cnt == n&&is_prime[a[]+a[n-]])
     {
         for(int i = ; i < n- ;i++)
         {
             printf("%d ",a[i]);
         }
         printf("%d\n",a[n-]);
     }

     for(int i =  ;i <= n ;i++)
     {
         if(!vis[i]&&is_prime[a[cnt-]+i])
         {
             a[cnt] = i;
             vis[i] = ;
             dfs(n,cnt+);
             vis[i] = ;
         }
     }
 }

 int main()
 {
     int n;
     int c = ;
     init();
     while(~scanf("%d",&n))
     {
         printf("Case %d:\n",++c);
         a[] = ;
         memset(vis,,sizeof(vis));
         vis[] = ;
         dfs(n,);
         puts("");
     }
     return ;
 }