A Knight's Journey
Time Limit: 1000MS?? Memory Limit: 65536K
Total Submissions: 24840?? Accepted: 8412

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 3Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4
?Source

简单的深搜，不多说，直接上代码！

 #include 
#include 
#include 
using namespace std; 
int pathlow[30],pathdown[30],visit[30][30]; 
int dir[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}},n,m;//这里注意是字典序最小 
bool dfs(int low,int down,int num) 
{ 
    int i,x,y; 
    if(num==n*m) 
    { 
     
        for(i=0;i=0&&x=0&&y
#include
#include
using namespace std;
int pathlow[30],pathdown[30],visit[30][30];
int dir[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}},n,m;//这里注意是字典序最小
bool dfs(int low,int down,int num)
{
 int i,x,y;
 if(num==n*m)
 {
 
  for(i=0;i=0&&x=0&&y 
 
?