A Knight's Journey
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 21849   Accepted: 7379

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

思路：给出一个国际棋盘的大小，判断马能否不重复的走过所有格，并记录下其中按字典序排列的第一种路径。

"马的遍历"是一道经典回溯题，当然还是DFS...这题有3个要密切注意的地方：

1、题目要求以"lexicographically"方式输出，也就是字典序...一开始没看懂这个词结果WA了N次...要以字典序输出路径，那么方向数组就要以特殊的顺序排列了...这样只要每次从dfs(1,1)开始搜索，第一个成功遍历的路径一定是以字典序排列...

2、国际象棋的棋盘，横行为字母，表示横行坐标的是y；纵行为数字，表示纵行的坐标是x...一开始又搞反了...

3、虽然题目没说，但是这道题最后一组数据后是没有空行的...否则会PE...




代码：

#include <stdio.h>
#include <stdlib.h>
#define MAX 27
int map[MAX][MAX], sx[MAX], sy[MAX];
int dir[8][2] = {{-2, -1}, {-2, 1}, {-1, -2}, {-1, 2},{1, -2}, {1, 2}, {2, -1}, {2, 1}};
int p, q, sign, step;

void dfs(int i, int j)
{
if (sign) 
 

return;
    

int x, y, k;
    

step++;
    

sx[step] = i;

    


sy[step] = j;

    


if(step == p * q)

    


{

        


sign = 1;

        


return;

    


}

    


map[i][j] = 1;

    

for (k = 0; k < 8; k++)
    

{
        

y = j + dir[k][0];

       


 x = i + dir[k][1];

        


if (map[x][y] == 0 && x > 0 && x <= p && y > 0 && y <= q)
        

{
    
            

dfs(x, y);
            

step--;

        


}
    

}
   

 map[i][j] = 0;



}
 




int main()


{
    

int i, j, n, t = 0;
   

 scanf("%d", &n);
    

while(n--)
    

{
        

sign = 0;
        

step = 0;
        

t++;
        

scanf("%d%d", &p, &q);
        

for (i = 1; i <= p; i++)
        

{
            

for(j = 1; j <= q; j++)
            

{
               

 map[i][j] = 0;
            

}
        

}
        

dfs(1, 1);

        

printf("Scenario #%d:\n", t);
        

if (sign)
        

{
            

for (i = 1; i <= p * q; i++)
           

 {
               

 printf("%c%d", sy[i] + 64, sx[i]);
            

}
           

 printf("\n");
        

}
       

 else
        

{
            

printf("impossible\n");
       

 }
        

if (n != 0) printf("\n");

    


}
    

//system("pause");
    

return 0;


}

链接：http://poj.org/problem?id=2488