题目：

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

31 12 34 3

Sample Output

Scenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4 题意： 给你一个p*q的棋盘，跳马在上面的任意一格开始移动，只能走‘日’字，问你能不能经过棋盘上面所有的格子；（百度的题意，明明是说经过所有的格子，有道硬是翻译 成了“找到一条这样的路，骑士每一次都要去一次”），输出要按照字典顺序输出 分析： 深度优先搜索，要经过所有的格子，那就肯定经过（1，1），所以就可以从（1，1）开始搜索； AC代码：

#include<iostream>

#include<cstring>

#include<cstdio>

using namespace std;

int t,p,q,flag;

int a[30][30];

int step[30][30];

int f[8][2]={

{1,-2},{-1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};

void dfs(int x,int y,int z)

{

    step[z][1]=x;

    step[z][2]=y;

     if (z==p*q)

     {

            flag=1;

            return ;

     }

    for (int i=0;i<8;i++)

        {

            int xi=x+f[i][0];

            int yi=y+f[i][1];

            if (xi>=1&&xi<=p&&yi>=1&&yi<=q&&!a[xi][yi]&&!flag)

            {

                a[xi][yi]=1;

                dfs(xi,yi,z+1);

                a[xi][yi]=0;

            }

        }

}

int main()

{

    cin>>t;

   for (int i=1;i<=t;i++)

    {

        flag=0;

       scanf("%d%d",&p,&q);

        memset(a,0,sizeof(a));

        memset(step,0,sizeof(step));

        a[1][1]=1;

        dfs(1,1,1);

        printf("Scenario #%d:\n",i);

        if (flag==1)

         {

             for (int j=1;j<=p*q;j++)

           printf("%c%d",step[j][2]+'A'-1,step[j][1]);

             printf("\n");

         }

        else

            printf("impossible\n");

        if (i!=t)

            printf("\n");

    }

    return 0;

}