HDU 1016 Prime Ring Problem (DFS)

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31031 Accepted Submission(s): 13755

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

简单DFS，给新生讲课用，题目要求顺时针和逆时针输出，实际好像不需要考虑这一点，最后的情况刚好是这样，素数判断用平方根内有无能整除此数的数字来判断。

 1 #include <iostream>
 2 #include <cmath>
 3 using    namespace    std;
 4 
 5 int    N;
 6 int    ANS[25];
 7 bool    VIS[25];
 8 
 9 bool    isprimer(int n);
10 void    dfs(int len);
11 int    main(void)
12 {
13     int    count = 0;
14 
15     while(cin >> N)
16     {
17         count ++;
18         fill(VIS,VIS + 22,false);
19         cout << "Case " << count << ":" << endl;
20         dfs(0);
21         cout << endl;
22     }
23 
24     return    0;
25 }
26 
27 void    dfs(int len)
28 {
29     if(len == N && isprimer(ANS[len - 1] + ANS[0]))
30     {
31         for(int i = 0;i < N - 1;i ++)
32             cout << ANS[i] << ' ';
33         cout << ANS[N - 1];
34         cout << endl;
35         return    ;
36     }
37 
38     for(int i = 1;i <= N;i ++)
39     {
40         if(!len && i > 1)
41             return;
42         if(len && !VIS[i])
43         {
44             if(isprimer(i + ANS[len - 1]))
45             {
46                 ANS[len] = i;
47                 VIS[i] = true;
48                 dfs(len + 1);
49                 VIS[i] = false;
50             }
51         }
52         else    if(!len)
53         {
54             ANS[len] = i;
55             VIS[i] = true;
56             dfs(len + 1);
57             VIS[i] = false;
58         }
59     }
60 
61 }
62 
63 bool    isprimer(int n)
64 {
65     for(int i = 2;i <= sqrt(n);i ++)
66         if(n % i == 0)
67             return    false;
68     return    true;
69 }