/*
A Knight's Journey
Time Limit: 1000MS Memory Limit: 65536K
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
*/
/*
本题为POJ2488 ,之前A过一次所以再写一遍只用了1个小时左右
大概意思为棋盘上的马走日..
能遍历棋盘就输出路线
用到了DFS回溯的思想,注意输出是字典序输出,也就是每走一步都从左下角开始优先遍历
*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int dicy[8] = {-2,-2,-1,-1,1,1,2,2};
int dicx[8] = {-1,1,-2,2,-2,2,-1,1}; //字典树排序 注意优先顺序
int bingo = 0;
int visit[100][100];
int trackx[200],tracky[200];
int x,y;
int length;
int ok(int n,int m) //判断是否在棋盘界内
{
if(n <= x && n >= 1 && m >= 1 && m <= y)
return 1;
else
return 0;
}
void dfs(int m,int n,int length) //深搜
{
trackx[length] = m;
tracky[length] = n;
if(length == x*y)
{
bingo = 1;
for(int j = 1 ; j <= length ; j++) //输出
{
printf("%c%d",tracky[j]+'A'-1,trackx[j]);
}
printf("\n\n");
}
for(int i = 0 ; i < 8 ; i++)
{
if(ok(m+dicx[i],n+dicy[i]) && !visit[m+dicx[i]][n+dicy[i]] && !bingo)
{
visit[m+dicx[i]][n+dicy[i]] = 1;
dfs(m+dicx[i],n+dicy[i],length+1);
//visit[m+dicx[i]][n+dicy[i]] = 0; //回溯
}
}
}
int main()
{
int n;
scanf("%d",&n);
for(int k = 1 ; k <= n ; k++)
{
bingo = 0;
for(int m = 0 ; m < 20 ; m++)
{
for(int n = 0 ; n < 20 ; n++)
{
visit[m][n] = 0 ;
}
}
scanf("%d%d",&x,&y); // 输入x,y x为列数,y为行数
trackx[1] = 1;
tracky[1] = 1;
visit[1][1] = 1;
printf("Scenario #%d:\n",k);
dfs(1,1,1);
if(!bingo)
{
printf("impossible\n\n");
}
}
return 0;
}
