hdu 1016 Prime Ring Problem（深度优先搜索）

Posted 2020-06-12

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；

 1 #include<iostream>
 2 #include<cstring>
 3 #define N 25
 4 #define M 40
 5 using namespace std;
 6 
 7 bool is_prime[M],visited[N];
 8 int n,test,ans[N];
 9 
10 void work(int k)
11 {
12     int i;
13     if(k==n+1)
14     {
15         if(!is_prime[ans[n]+ans[1]]) return ;
16         for(i=1;i<=n-1;i++)
17             cout<<ans[i]<<" ";
18         cout<<ans[i]<<endl;
19         return ;
20     }
21     for(i=2;i<=n;i++)
22     {
23         if(!visited[i]&&is_prime[ans[k-1]+i])
24         {
25             visited[i]=true;
26             ans[k]=i;
27             work(k+1);
28             visited[i]=false;
29         }
30     }
31 }
32 
33 bool prime(int n)
34 {
35     if(n==1) return false;
36     if(n==2||n==3) return true;
37     int i;
38     for(i=2;i<n;i++)
39         if(n%i==0)
40             return false;
41     return true;
42 }
43 
44 int main()
45 {
46     int i;test=1;
47     for(i=1;i<M;i++) is_prime[i]=prime(i);
48     while(cin>>n)
49     {
50         ans[1]=1;
51         memset(visited,false,sizeof(visited));
52         cout<<"Case "<<test<<":"<<endl;
53         work(2);
54         test++;
55         cout<<endl;
56     }
57     return 0;
58 }

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