POJ2488 A Knight's Journey（DFS+字典序输出）

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：按照国际象棋中马的跳法判断能否在所给棋盘里把所有格子遍历一遍，如果能则输出字典序最小的顺序。

解题思路：（神TM字典序最小！！！这个坑我给跪了Orz）

如果能够遍历图中所有的点，那么不管哪个顺序的起点一定是A1。（想一下为什么）

那么剩下的就是DFS了，如果能走完所有格子即判断能否走过n*m个格子。

最需要注意的就是搜索的方向顺序！（自己根据自己的坐标系画一下）

我的是x,tx,n,row相对应，y,ty,m,line相对应，这时的搜索顺序应该是

nx[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}}

AC代码：

#include<stdio.h>
#include<string.h>

int line[200],row[200],book[200][200];
int n,m,flag,nx[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
char str[200];

void dfs(int x,int y,int cnt)
{
	if(flag) return;
	if(cnt==n*m)
	{
		printf("A1");
		for(int i=1;i<cnt;i++)
			printf("%c%d",str[line[i]],row[i]);
		printf("\n");
		flag=1;
	}
	else
	{
		for(int i=0;i<8;i++)
		{
			int tx=x+nx[i][0];
			int ty=y+nx[i][1];
			if(tx>=1&&tx<=n&&ty>=1&&ty<=m&&book[tx][ty]==0)
			{
				row[cnt]=tx;
				line[cnt]=ty;
				book[tx][ty]=1;
				dfs(tx,ty,cnt+1);
				if(flag) return;
				book[tx][ty]=0;
			}
		}
	}
}

int main()
{
	int t,tt=1;
	char c;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		for(int i=1,c='A';i<=m;i++)
			str[i]=c++;
		flag=0;
		memset(book,0,sizeof(book));
		printf("Scenario #%d:\n",tt++);
		book[1][1]=1;
		dfs(1,1,1);
		if(flag==0) printf("impossible\n");
		printf("\n");
	}
	return 0;
}