国际象棋

/*                          A Knight's Journey
Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other

Problem 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

题意：问在给的p*q的棋盘上，以象棋中马的行走方式，是否可以不重复的走完整个棋盘。
若可以则输出任意一种行走方式，注意要按照字典序的方式输出。不可以就输出“impossible”。*/
#include <stdio.h>
#include<string.h>
char  map[60];
int book[27][27];
int flag;
int n,m,w=0;
int count;
int dx[8]={-2,-2,-1,-1,1,1,2,2};
int dy[8]={-1,1,-2,2,-2,2,-1,1};
void dfs(int x,int y)
{
int i,j,k;
int sx,sy;
if(flag)return;
if(count==n*m)
{
flag=1;
for(i=0;i<count*2;i++)
{
printf("%c",map[i]);

}
printf("\n\n");
}
if(x>=0&&y>=0&&x<n&&y<m&&!book[x][y])
{
map[w]='A'+y;//y代表的行
map[w+1]='1'+x;//x代表的列
count++;
w+=2;
book[x][y]=1;
for(k=0;k<8;k++)
{
sx=dx[k]+x;
sy=dy[k]+y;
dfs(sx,sy);
}
count--;
book[x][y]=0;
w-=2;
}
}
int main()
{
int t,r=0;
scanf("%d",&t);
while(t--)
{
memset(book,0,sizeof(book));
scanf("%d%d",&n,&m); // n代表列  m代表行
flag=0;
count=0;
printf("Scenario #%d\n",++r);
dfs(0,0);
if(!flag)
printf("impossible\n");
}
return 0;

}