OpenJudge-Mooc 2:A Knight's Journey(dfs)

总时间限制: 

1000ms 

内存限制: 

65536kB

描述

Background
The knight is getting bored of seeing the same black and white squares again and 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.

输入

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 many different square numbers 1, . . . , p exist, q describes how many different square letters exist. These are the first q letters of the Latin alphabet: A, . . .

输出

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 should 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 should output impossible on a single line.

样例输入

31 12 34 3

样例输出

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

/* * dfs、 注意要按照字典序来设置dir[8][2]..*/#include "iostream"#include "cstring"using namespace std;int col, row;int pathI[30], pathJ[30];bool visited[27][27];int dir[8][2] = { { -2,-1 },{ -2,1 },{ -1,-2 },{ -1,2 },{ 1,-2 },{ 1,2 },{ 2,-1 },{ 2,1 } };bool dfs(int p, int q, int step) {pathI[step] = p;pathJ[step] = q;if (step == col*row) {return true;}for (int i = 0; i < 8; i++) {int k = p + dir[i][0];int l = q + dir[i][1];if ((!visited[k][l]) && k >= 0 && k<row && l >= 0 && l<col) {visited[k][l] = 1;if (dfs(k, l, step + 1))return true;visited[k][l] = 0;}}return false;}int main() {int n;cin >> n;for (int i = 1; i <= n; i++) {cin >> col >> row;for (int i = 0; i <= 27; i++)for (int j = 0; j <= 27; j++)visited[i][j] = 0;visited[0][0] = 1;cout << "Scenario #" << i << ":" << endl;if (dfs(0, 0, 1))for (int i = 1; i <= row*col; i++)printf("%c%d", pathI[i] + 'A', pathJ[i] + 1);elsecout << "impossible";cout << endl;if (i != n)cout << endl;}return 0;}