poj 2488 A Knight's Journey

题意：给你一个n*m的矩阵，看能从（1,1）点开始按照字典序开始遍历，能否遍历n*m内的所有的点
盗张图～～ 没找到，但我见到过，、等找到再发吧～～
按照字典序遍历，其实就是规定遍历的优先方向，也就是下面这个

int Dir[][2] = {{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <set>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#define LL long long
#define INF 0x3f3f3f3f
#define RR freopen("in.txt","r",stdin)
//#pragma comment(linker,"/SATACK,;1024000,1024000")
#define MAX 1100
using namespace std;
int Dir[][2] = {{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};
bool vis[MAX][MAX];
struct node
{
int row;
char cal;
} way[MAX];
int p,q;
bool flag;
int Max_step;

void DFS(int x,int y,int step)
{
if(step == Max_step)
{
for(int i=1; i<=Max_step; i++)
{
printf("%c%d",way[i].cal,way[i].row);
}
printf("\n");
flag = true;
return ;
}

for(int i=0; i<8 && !flag ; i++)
{
int xx = x + Dir[i][0];
int yy = y + Dir[i][1];
if(!vis[xx][yy] && xx > 0 && yy > 0 && xx <= q && yy <= p)
{
way[step+1].cal = 'A' + xx - 1;
way[step+1].row = yy;
vis[xx][yy] = true;
DFS(xx,yy,step+1);
vis[xx][yy] = false;
}
}

}
int main()
{
// RR;
int n;
int t = 0;
cin>>n;
while(n--)
{
cin>>p>>q;
printf("Scenario #%d:\n",++t);
Max_step = p * q;
memset(vis,false,sizeof(vis));
flag = false;
way[1].cal = 'A';
way[1].row = 1;
vis[1][1] = true;
DFS(1,1,1);
if(!flag)
printf("impossible\n");
printf("\n");
}
return 0;
}