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.

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, . . .

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.

Scenario #1:

A1

Scenario #2:

impossible

Scenario #3:

A1B3C1A2B4C2A3B1C3A4B2C4

经典的DFS，大概题意就是让骑士走遍棋盘的所有格子。一路走到黑，看这条路能不能使骑士把格子都走完。

 #include<cstdio>

 #include<string.h>

 using namespace std;

 int vis[][];

 char str[];

 int s[][]={-,-,-,,-,-,-,,,-,,,,-,,};//这种处理方式很常用的，要记住，即骑士当前所在位置的周围所能走到的位置

 int p,q;

 int dfs(int x,int y,int sum,int cnt)

 {

     if(sum==p*q) return ;//sum是记录走过的格子

     int x1,y1;

     for(int i=;i<;i++)

     {

         x1=x+s[i][];

         y1=y+s[i][];

         if(x1>=&&x1<q&&y1>=&&y1<p&&!vis[x1][y1])//边界条件及判断是否访问过吗

         {

             vis[x1][y1]=;

             str[cnt+]=x1+'A';//开一个str数组，用来记录

             str[cnt+]=y1+'';

             if(dfs(x1,y1,sum+,cnt+))//记得这里要写成sum+1,cnt+2

                 return ;

             vis[x1][y1]=;//回溯

         }

     }

     return ;

 }

 int main()

 {

     int t;

     scanf("%d",&t);

     for(int i=;i<=t;i++)

     {

         scanf("%d %d",&p,&q);

         memset(vis,,sizeof(vis));

         memset(str,,sizeof(str));

         str[]='A';

         str[]='';

         vis[][]=;

         if(dfs(,,,)){

             printf("Scenario #%d:\n",i);

             for(int j=;j<strlen(str);j++)

                 printf("%c",str[j]);

             printf("\n\n");//记得每组数据输出后，有个空行

         }

         else{

             printf("Scenario #%d:\n",i);

             printf("impossible\n\n");

         }

     }

     return ;

 }