poj2488 A Knight's Journey

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 3

Sample 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<iostream>
#include<stdio.h>
#include<cstring>
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<n*m;i++)
		{
		
			printf("%c%d",'A'+pathdown[i],pathlow[i]+1);
		}
		return true;
	}

	for(i=0;i<8;i++)
	{
		x=low+dir[i][0];
		y=down+dir[i][1];
		pathlow[num]=x;
		pathdown[num]=y;
		if(x>=0&&x<n&&y>=0&&y<m&&(!visit[x][y]))
		{
		
			visit[x][y]=1;
			if(dfs(x,y,num+1))
			{
			
				return true;
			}
			else
			{
			
				visit[x][y]=0;//这里要注意，一定要重新标记为0
			}
		}

	}
	return false;

}
int main ()
{
	int t,i;
	while(scanf("%d",&t)!=EOF)
	{
	
		for(i=1;i<=t;i++)
		{
			printf("Scenario #%d:
",i);
			scanf("%d%d",&n,&m);
			memset(visit,0,sizeof(visit));
			visit[0][0]=1;
			pathlow[0]=0;
			pathdown[0]=0;
			if(!dfs(0,0,1))
			{
			
				printf("impossible");
			}
			printf("

");


		}
	}

	return 0;
}