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.

3

1 1

2 3

4 3

Scenario #1:

A1

Scenario #2:

impossible

Scenario #3:

A1B3C1A2B4C2A3B1C3A4B2C4

 #include <cstdio>

 #include <iostream>

 #include <string>

 #include <cstring>

 //lexicographically first path

 //题目output中出现了以上几个单词

 //就是说骑士的移动步子是按照字典顺序来排列的

 //首先对列进行排序较小的在前面，然后按行进行排序也是较小的在前面

 //我构造的xy坐标是按照SDL中的图形坐标，x往下递增，y往右递增

 //转换成数学上的xy坐标就要先对列x做排序然后再对行y做排序

 int dirx[]={-,-,-,-,,,,};

 int diry[]={-,,-,,-,,-,};

 int T,p,q,mark[][];

 std::string way;

 bool DFS(int x,int y,int step)

 {

    if(step==p*q)

       {

          return true;

       }

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

       {

          int dx=x+dirx[i];

          int dy=y+diry[i];

          if((dx>= && dx<=q) && (dy>= && dy<=p) && mark[dx][dy]!=)

             {

                mark[dx][dy]=;

                //当找到路径才会插入操作，所以在main函数循环中不需要清空string

                if(DFS(dx,dy,step+))

                   {

                      //在第0个位置插入1个字符

                      //先插入数字然后再插入字母，否则顺序颠倒

                      way.insert(,,dy+'');

                      way.insert(,,dx+'A'-);

                      return true;

                   }

                mark[dx][dy]=;

             }

       }

    return false;

 }

 int main()

 {

    scanf("%d",&T);

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

       {

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

          way.clear();//每次都要清空string

          bool find=false;

          for(int x=;x<=q && !find;x++)

             {

                for(int y=;y<=p;y++)

                   {

                      memset(mark,,sizeof(mark));

                      mark[x][y]=;

                      if(DFS(x,y,))

                         {

                            //找到的话插入起始位置

                            way.insert(,,y+'');

                            way.insert(,,x+'A'-);

                            find=true;

                            break;

                         }

                   }

             }

          std::cout<<"Scenario #"<<i<<":"<<std::endl;

          if(find)

             {

                //整体输出就行

                std::cout<<way<<std::endl<<std::endl;

             }

          else

             printf("impossible\n\n");

       }

    return ;

 }