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

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本文探讨了骑士周游问题的解决方法，这是一种经典的回溯算法应用案例。问题要求骑士在棋盘上行走，每格仅能经过一次，并最终回到起点。文章提供了一种基于深度优先搜索（DFS）的解决方案，通过设定不同的方向移动顺序确保了解的字典序最小。

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

样例输出

Scenario #1: A1 Scenario #2: impossible Scenario #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);
		else
			cout << "impossible";
		cout << endl;
		if (i != n)
			cout << endl;
	}
	return 0;
}