Poj 2488 A Knight & amp; #39; s Journey (DFS)

Poj 2488 A Knight & #39; s Journey (DFS)

 

A Knight's Journey

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:34660		Accepted:11827

 

Description

Background
The knight is getting bored of seeing the same black and white squares again and has decided to make a journey
Around the world. whenever a knight moves, it is two squares in one direction and one square perpendicular to this. the world of a knight is the chessboard he is living on. our knight lives on a chessboard that has a smaller area than a regular 8*8 board, but it is still rectangular. can you help this adventurous knight to make travel plans?

Problem
Find a path such that the knight visits every square once. The knight can start and end on any square of the board.

Input

The input begins with a positive integer n in the first line. the following lines contain n test cases. each test case consists of a single line with two positive integers p and q, such that 1 <= p * q <= 26. this represents a p * q chessboard, where p describes how between different square numbers 1 ,..., p exist, q describes how many different square letters exist. these are the first q letters of the Latin alphabet: ,...

Output

The output for every scenario begins with a line containing Scenario # I:, where I is the number of the scenario starting at 1. then print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves followed by an empty line. the path shoshould be given on a single line by concatenating the names of the visited squares. each square name consists of a capital letter followed by a number.
If no such path exist, you shoshould output impossible on a single line.

Sample Input

31 12 34 3

Sample Output

Scenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4

Source

TUD Programming Contest 2005, Darmstadt, Germany

 

Question link: http://poj.org/problem? Id = 2488

 

A checkerboard with n * m grids, simulating a horse taking the word "day", and finding out whether all the grids of a checkerboard can be completed. If yes, find the path, that is, it follows the order of the lattice. If there are multiple cases, the smallest case is in the Lexicographic Order. The grid position is indicated by Y-axis. the y-axis is a letter and the y-axis is a number.

 

Solution: DFS is searched from the first grid, because each step can only take the word "day", so each step has eight directions, because the smallest Lexicographic Order is followed, and the enumeration direction is also alphabetically. Mark the path in the return process of the solution.

 

The Code is as follows:

 

# Include
 
  
# Include
  
   
Int dx [8] = {-2,-2,-1,-1, 1, 2 }; // enumerate eight int dy [8] = {-,-} in the Lexicographic Order; int xx [26] = {'A ', 'B', 'C', 'D', 'E', 'F', 'G', 'h', 'I', 'J', 'k ', 'l', 'M', 'n', 'O', 'P', 'Q', 'R', 's', 't', 'U ', 'V', 'w', 'x', 'y', 'z'}; char ansx [28]; // record the abscissa path int ansy [28]; // record the ordinate path int sum, m, n, j; bool p; int a [28] [28], vis [28] [28]; void dfs (int x, int y, int cnt) {if (cnt = sum) // all the grids have been completed {ansx [j] = xx [x]; ansy [j ++] = y + 1; p = true; return ;}for (int I = 0; I <8; I + +) {If (x + dx [I] <0 | x + dx [I]> = n | y + dy [I] <0 | y + dy [I]> = m) continue; if (vis [x + dx [I] [y + dy [I]) continue; vis [x + dx [I] [y + dy [I] = 1; dfs (x + dx [I], y + dy [I], cnt + 1); if (p) // if the grid is successfully completed, mark the path of each step forward {ansx [j] = xx [x]; ansy [j ++] = y + 1; return;} vis [x + dx [I] [y + dy [I] = 0 ;}} int main () {int t; scanf (% d, & t); for (int cnt = 1; cnt <= t; cnt ++) {p = false, j = 0; memset (vis, 0, sizeof (vis); vis [0] [0] = 1; scanf (% d, & m, & n ); sum = n * m; dfs (0, 0, 1); printf (Scenario # % d:, c Nt); if (! P) printf (impossible); else // output path {for (int I = J-1; I> = 0; I --) printf (% c % d, ansx [I], ansy [I]); printf ();} if (cnt