B-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

给一个n1*n2的棋盘，从(0，0)出发，每次走日字形，能否不重复的遍历所有的点.从A1这个格子开始搜索，就能保证字典序是小的；除了这个条件，我们还要控制好马每次移动的方向，控制方向时保证字典序最小（即按照下图中格子的序号搜索）。控制好这两个条件，直接从A1开始深搜就行了。

#include<cstdio>  #include<cstring>  #include<algorithm>  using namespace std;  int path[88][88], vis[88][88], p, q, cnt;  bool flag;  int dx[8] = {-1, 1, -2, 2, -2, 2, -1, 1};  int dy[8] = {-2, -2, -1, -1, 1, 1, 2, 2};  bool judge(int x, int y)  {      if(x >= 1 && x <= p && y >= 1 && y <= q && !vis[x][y] && !flag)          return true;      return false;  }  void DFS(int r, int c, int step)  {      path[step][0] = r;      path[step][1] = c;      if(step == p * q)      {          flag = true;          return ;      }      for(int i = 0; i < 8; i++)      {          int nx = r + dx[i];          int ny = c + dy[i];          if(judge(nx,ny))          {              vis[nx][ny] = 1;              DFS(nx,ny,step+1);              vis[nx][ny] = 0;          }      }  }  int main()  {      int i, j, n, cas = 0;      scanf("%d",&n);      while(n--)      {          flag = 0;          scanf("%d%d",&p,&q);          memset(vis,0,sizeof(vis));          vis[1][1] = 1;          DFS(1,1,1);          printf("Scenario #%d:\n",++cas);          if(flag)          {              for(i = 1; i <= p * q; i++)                  printf("%c%d",path[i][1] - 1 + 'A',path[i][0]);          }          else              printf("impossible");          printf("\n");          //if(n != 0)              printf("\n");      }      return 0;  }