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
题目大意：
给出一个p行q列的国际棋盘，马可以从任意一个格子开始走，问马能否不重复的走完所有的棋盘。如果可以，输出按字典序排列出最小的路径。打印路径时，列用大写字母表示（A表示第一列），行用阿拉伯数字表示（从1开始），先输出列，再输出行。
分析：如果马可以不重复的走完所有的棋盘，那么它一定可以走到A1这个格子。所以我们只需从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开始搜索就行了。
程序代码：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int map[101][101],vis[101][101],p,q,ans,flag;
int next1[8]={-1, 1,-2, 2,-2,2,-1,1};//分别为图中的八个点
int next2[8]={-2,-2,-1,-1, 1,1, 2,2};//按照字典序定义每个点
bool judge(int x,int y)
{
if(x>=1&&x<=p&&y>=1&&y<=q&&!vis[x][y]&&!flag)
return true;//在地图范围内，这个点没有走过，并且没有走完地图
return false;
}
void dfs(int x,int y,int step)
{
map[step][0]=x;//横坐标
map[step][1]=y;//纵坐标
if(step==p*q)//此时已经走完整个地图
{
flag=1;
return ;
}
for(int i=0;i<8;i++)//八个方向去搜索吧！！！
{
int tx=x+next1[i];
int ty=y+next2[i];
if(judge(tx,ty))
{
vis[tx][ty]=1;
dfs(tx,ty,step+1);
vis[tx][ty]=0;
}
}
}
int main()
{
int n,ans=0;
cin>>n;
while(n--)
{
flag=0;
cin>>p>>q;
memset(vis,0,sizeof(vis));
vis[1][1]=1;
dfs(1,1,1);
cout<<"Scenario #"<<++ans<<":"<<endl;
if(flag)
{
for(int i=1;i<=p*q;i++)
printf("%c%d",map[i][1]-1+'A',map[i][0]);
} //因为棋盘横纵坐标最大就是3，所以这个数减1再加上‘A’的ASCII码就可以了
else
cout<<"impossible";
cout<<endl;
if(n)//注意题目的输出格式中的换行，避免PE
cout<<endl;
}
return 0;
}