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 }