POJ2488 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。目标是在一个缩小版的国际象棋棋盘上找到一条路径，使得骑士能够访问每一个方格恰好一次。文章提供了详细的算法实现代码，并通过样例输入输出展示了如何确定可行路径。

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Description

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.

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 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, . . .

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 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.

Sample Input

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

国际象棋中马的走法，比较棘手的是题目要求输出路径，并且是按照字典序，故需要从第一个格开始。

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
struct node
{
	/* data */
	int row;
	char col;
};
int p, q, x, y;
bool map['Z' + 1][27];
void path(int i, int j, int num)
{
	switch(num) {
		case 1: {
			x = i - 1;
			y = j - 2;
			break;
		}
		case 2: {
			x = i + 1;
			y = j - 2;
			break;
		}
		case 3: {
			x = i - 2;
			y = j - 1;
			break;
		}
		case 4: {
			x = i + 2;
			y = j - 1;
			break;
		}
		case 5: {
			x = i - 2;
			y = j + 1;
			break;
		}
		case 6: {
			x = i + 2;
			y = j + 1;
			break;
		}
		case 7: {
			x = i - 1;
			y = j + 2;
			break;
		}
		case 8: {
			x = i + 1;
			y = j + 2;
		}
	}
}
bool dfs(node* a, int i, int j, int step) 
{
	map[i][j] = true;
	a[step].row = i;
	a[step].col = j;
	if(step == a[0].row) return true;
	for(int k = 1; k <= 8; ++k) {
		path(i, j, k);
		int ii = x, jj = y;
		if(!map[ii][jj] && ii >= 1 && ii <= p && jj >= 'A' && jj <= 'A' + q - 1)
			if(dfs(a, ii, jj, step + 1)) return true;
	}
	map[i][j] = false;
	return false;
}
int main(int argc, char const *argv[])
{
	int n;
	cin >> n;
	int t = 1;
	while(t <= n) {
		memset(map, false, sizeof(map));
		cin >> p >> q;
		if(p == 1 && q == 1) {
			cout << "Scenario #" << t++ << ":" << endl;
			cout << "A1" << endl << endl;
			continue;
		}
		node a[p * q + 1];
		a[0].row = p * q;
		bool flag = false;
		for(int j = 'A'; j <= 'A' + q - 1; ++j) {
			for(int i = 1; i <= p; ++i)
				if(dfs(a, i, j, 1)) {
					cout << "Scenario #" << t++ << ":" << endl;
					for(int k = 1; k <= a[0].row; ++k)
						cout << a[k].col << a[k].row;
					cout << endl << endl;
					flag = true;
					break;
				}
			if(flag) break;
		}
		if(!flag) {
			cout << "Scenario #" << t++ << ":" << endl;
			cout << "impossible" << endl << endl;
		}
	}
	return 0;
}