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。

 #include <iostream>
 #include <cmath>
 using    namespace    std;

 int    N;
 int    ANS[];
 bool    VIS[];

 bool    isprimer(int n);
 void    dfs(int len);
 int    main(void)
 {
     int    count = ;

     while(cin >> N)
     {
         count ++;
         fill(VIS,VIS + ,false);
         cout << "Case " << count << ":" << endl;
         dfs();
         cout << endl;
     }

     return    ;
 }

 void    dfs(int len)
 {
     if(len == N && isprimer(ANS[len - ] + ANS[]))
     {
         for(int i = ;i < N - ;i ++)
             cout << ANS[i] << ' ';
         cout << ANS[N - ];
         cout << endl;
         return    ;
     }

     for(int i = ;i <= N;i ++)
     {
         if(!len && i > )
             return;
         if(len && !VIS[i])
         {
             if(isprimer(i + ANS[len - ]))
             {
                 ANS[len] = i;
                 VIS[i] = true;
                 dfs(len + );
                 VIS[i] = false;
             }
         }
         else    if(!len)
         {
             ANS[len] = i;
             VIS[i] = true;
             dfs(len + );
             VIS[i] = false;
         }
     }

 }

 bool    isprimer(int n)
 {
     for(int i = ;i <= sqrt(n);i ++)
         if(n % i == )
             return    false;
     return    true;
 }