（不会下象棋的我只会背马走日）A Knight's Journey

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

3
1 1
2 3
4 3
Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

 

*/马走日，象走田；
 车走直路炮翻山；
 士走斜线护将边；
 小卒一去不回还。*/
#include<iostream>
#include<string.h>
using namespace std;
int mapp[27][27], book[27][27];//mapp记录路径，book记录位置是否被访问，我用map竟然会报错。。。。
int flag,p,q;//p,q一定要定义全局变量
int nexxt[8][2] = { {-1,-2},{1,-2},{-2,-1},{2,-1},
	{-2,1},{2,1},{-1,2},{1,2} };//马走日，8个方向
void dfs(int x, int y, int step)
{//关键所在！！！最小遍历路径按字典序排序肯定是从第一个点开始的，因为第一个点为A1，字典序最小
	mapp[step][0] = x;//存储路径，这个是数字坐标
	mapp[step][1] = y;//这个是字母坐标
	//可以拿笔模拟一下，就清楚明白了（滑稽）
	if (step == p * q)
	{//遍历完成
		flag = 1;
		return;
	}
	for (int k = 0; k <= 7; k++)
	{//从8个方向寻找
		int tx = x + nexxt[k][0];
		int ty = y + nexxt[k][1];
		if (tx >= 1 && tx <= p && ty >= 1 && ty <= q && !book[tx][ty] && !flag)
		{//边界条件一定要找对，因为是从（1，1）开始的，得带=；没被标记过，没到终点！！！
			book[tx][ty] = 1;//标记该点走过；
			dfs(tx, ty, step + 1);//从该点开始，步数加一
			book[tx][ty] = 0;//取消标记，之前说过原因
		}
	}
}
int main()
{
	int sd, s = 1;
	cin >> sd;
	while (sd--)
	{//初始化标记量
		flag = 0;
		cin >> p >> q;
		memset(book, 0, sizeof(book));//每次初始化标记数组为0，未标记
		book[1][1] = 1;//标记，从第一个点开始
		dfs(1, 1, 1);//包括第一个点，步数为1>>>>搜索完毕
		printf("Scenario #%d:\n", s++);//注意输出样式
		if (flag)//如果遍历完毕，就输出路径
		{
			for (int i = 1; i <= p * q; i++)
			//关键所在，遍历整个地图，把前一个坐标转换为字母，后一个原样输出q
			//先输出字母坐标，后输出数字坐标！！！！
				printf("%c%d", mapp[i][1] - 1 + 'A', mapp[i][0]);
			//注意是从1开始的，要先减1再加'A'
			cout << endl << endl;//换两次行
		}
		else//不可能遍历
			cout << "impossible" << endl << endl;
	}
}
inline char gc()
{
	static char buff[100000000], *S = buff, *T = buff;
	return S == T && (T = (S = buff) + fread(buff, 1, 100000000, stdin), S == T) ? EOF : *S++;
}