POJ 2243 Knight Moves bfs

• A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves
  that visits each square of a given set of n squares on a chessboard
  exactly once. He thinks that the most difficult part of the problem
  is determining the smallest number of knight moves between two given
  squares and that, once you have accomplished this, finding the tour
  would be easy.
  Of course you know that it is vice versa. So you offer him to write a program that solves the “difficult” part.
  Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest
  route from a to b.

Input
The input will contain one or more test cases. Each test case consists of one line containing two squares separated by one space. A square is a string consisting of a letter (a-h) representing the column and a digit (1-8) representing the row on the chessboard.
Output
For each test case, print one line saying “To get from xx to yy takes n knight moves.”.
Sample Input

e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6

Sample Output

To get from e2 to e4 takes 2 knight moves.
To get from a1 to b2 takes 4 knight moves.
To get from b2 to c3 takes 2 knight moves.
To get from a1 to h8 takes 6 knight moves.
To get from a1 to h7 takes 5 knight moves.
To get from h8 to a1 takes 6 knight moves.
To get from b1 to c3 takes 1 knight moves.
To get from f6 to f6 takes 0 knight moves.

/*题意描述：给出两个字符串用空格隔开，字符串中第一个元素为字母第二个为数字，字母是（a_{h）数字是（1}8）。按照马走日的格式来走，然后问第一个最短需要走多少步才能走到第二个哪里。

解题思路：用字符串的形式输入，把它分离开来，把字母转换成（1-8）的数字，然后把两个字符串转换成坐标。 */

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

int e[10][10],book[10][10];
int a,b,c,d;
int min;
char s[5],s1[5];

int next[8][2]={2,1, 1,2, -2,1, -1,2, 2,-1, 1,-2, -2,-1, -1,-2};
struct node
{
	int x,y,s;
}que[110];

void bfs(int x,int y)
{
	int head,tail,tx,ty;
	int k;
	head=1;
	tail=1;
	que[tail].x=x;
	que[tail].y=y;
	que[tail].s=0;
	book[a][b]=1;
	tail++;
	int flag=0;
	if(a==c&&b==d)
		printf("To get from %s to %s takes 0 knight moves.\n",s,s1);
	else
	{
		while(head<tail)
		{
			for(k=0;k<8;k++)
			{
				tx=que[head].x+next[k][0];
				ty=que[head].y+next[k][1];
				
				if(tx<1||ty<1||tx>8||ty>8)
					continue;
				if(book[tx][ty]==0)
				{
					book[tx][ty]=1;
					que[tail].x=tx;
					que[tail].y=ty;
					que[tail].s=que[head].s+1;
					tail++;
				}
				if(tx==c&&ty==d)
				{
					flag=1;
					break;
				}
			}
			if(flag==1)
			{
				printf("To get from %s to %s takes %d knight moves.\n",s,s1,que[tail-1].s);
				break;
			}
			head++;
		}	
	}	
 }
 
 int main(void) 
 {
 	while(scanf("%s%s",s,s1)!=EOF)
 	{
 		memset(book,0,sizeof(book));
 		min=99999999;
 		a=s[0]-'a'+1;
 		b=s[1]-'0';
 		c=s1[0]-'a'+1;
 		d=s1[1]-'0';
 		bfs(a,b);
	 }
	 return 0;
 }