Hdu 1016 Prime Ring Problem (deep first search), hdu deep first search

Hdu 1016 Prime Ring Problem (deep first search), hdu deep first search
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 DescriptionA ring is compose of n circles as shown in digoal. put natural number 1, 2 ,..., n into each circle separately, and the sum of numbers in two adjacent circles shoshould be a prime.
Note: the number of first circle shoshould always be 1.
 

 

Inputn (0 <n <20 ).

 

OutputThe 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 abve process.
Print a blank line after each case.

 

Sample Input6 8

 

Sample OutputCase 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 6 1 7 6 5 8 3 2 1 6 7 4 3 8 5 2 input n to find 1 ~ The combination of n, so that the adjacent two numbers are the prime number. Analysis: preprocessing the prime number between 40, and then backtracking; 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}View Code