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): 37484    Accepted Submission(s): 16554


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
 
解题思路:
  简单的DFS即可
 1 #include <iostream>
 2 #include <cmath>
 3 #include <string.h>
 4 int n;
 5 int ans[20],used[20];
 6 using namespace std;
 7 bool isprime(int m)
 8 {
 9     for(int i=2;i<=sqrt(m*1.0);i++)
10     {
11         if(m%i==0)  return 0;
12     }
13     return 1;
14 }
15 void dfs(int x)
16 {
17     if(x==n+1 && isprime(ans[1]+ans[n]))
18     {
19         for(int i=1;i<n;i++)
20         {
21             cout<<ans[i]<<' ';
22         }
23         cout<<ans[n]<<endl;
24     }
25     else
26     {
27         for(int j=2;j<=n;j++)
28         {
29             if(used[j]==0 && isprime(j+ans[x-1]))
30             {
31                 ans[x]=j;
32                 used[j]=1;
33                 dfs(x+1);
34                 used[j]=0;
35             }
36         }
37     }
38 }
39 int main()
40 {
41     int k=1;
42     while(cin>>n)
43     {
44         memset(used,0,sizeof(used));
45         printf("Case %d:\n",k);
46         k++;
47         ans[1]=1;
48         used[1]=1;
49         dfs(2);
50         cout<<endl;
51     }
52 }

 

posted @ 2016-01-23 23:30  nicetomeetu  阅读(115)  评论(0编辑  收藏  举报