Poj 2488 -- a knight's journey

A knight's journey

Time limit:1000 ms		Memory limit:65536 K
Total submissions:30388		Accepted:10404

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 ining "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

Let the server guard traverse the entire graph in Lexicographic Order. Each vertex can only be reached once.

Train of Thought: the data is not big, just search directly. Note that each test data is followed by a blank line.

1/* ============================================== ====================================== 2 * Author: kevin 3 * filename: aknightsjourney. CPP 4 * creat time: 5 * description: 6 ================================================ ===================================== */7 # include <iostream> 8 # include <algorithm> 9 # include <cstdio> 10 # include <cstring> 11 # include <queue> 12 # include <cmath> 13 # define CLR (, b) Memset (a, B, sizeof (A) 14 # define M 3015 using namespace STD; 16 struct node17 {18 int X, Y; 19} steps [M * m]; 20 bool vis [m] [m]; 21 int p, q; 22 int dir [8] [2] = {-1,-2}, {1, -2}, {-2,-1}, {2,-1 }}; 23 int DFS (int x, int y, int CNT) 24 {25 if (vis [x] [Y]) {26 return-1; 27} 28 vis [x] [Y] = true; 29 steps [CNT]. X = x; steps [CNT]. y = y; 30 if (CNT = p * q-1) return 1; 31 for (INT I = 0; I <8; I ++) {32 in T xx = x + dir [I] [0]; 33 int YY = Y + dir [I] [1]; 34 if (XX> = 1 & XX <= P & YY> = 1 & YY <= q) {35 int ans = DFS (XX, YY, CNT + 1); 36 IF (ANS = 1) {37 return true; 38} 39 else if (ANS = 0) {40 vis [XX] [YY] = false; 41} 42} 43} 44 return 0; 45} 46 int main (INT argc, char * argv []) 47 {48 int N, T = 1; 49 scanf ("% d", & N); 50 while (n --) {51 CLR (steps, 0 ); 52 scanf ("% d", & P, & Q); 53 int CNT = 0; 54 printf ("Scenario # % d: \ n ", t ++ ); 55 int flag = 0; 56 for (INT I = 1; I <= Q; I ++) {57 for (Int J = 1; j <= P; j ++) {58 CLR (VIS, 0); 59 If (DFS (J, I, CNT) = 1) {// because I and j are reversed, contributed several times wr60 flag = 1; 61 break; 62} 63} 64 if (FLAG) break; 65} 66 If (! Flag) {67 printf ("impossible"); 68} 69 else {70 for (INT I = 0; I <Q * P; I ++) {71 printf ("% C % d", steps [I]. y-1 + 'A', steps [I]. x); 72} 73} 74 printf ("\ n"); 75} 76 return 0; 77}

View code