hdu 1372 骑士从起点走到终点的步数 (BFS)

给出起点和终点 求骑士从起点走到终点所需要的步数

Sample Input
e2 e4 //起点 终点
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.

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <queue>
 using namespace std;

 int map[][];
 int sx, sy;
 int ex, ey;
 bool flag;
 char s1[] , s2[] ;

 int dx[] = {-,-,,,-,-,,} ;
 int dy[] = {,,,,-,-,-,-} ;

 struct node
 {
     int x , y , step ;
 };

 int bfs()
 {
     queue<node> q ;
     int i , fx ,fy ;
     node now , t ;
     now.x = sx ;
     now.y = sy ;
     now.step =  ;
     q.push(now) ;
     memset(map ,  , sizeof(map)) ;
     map[sx][sy] =  ;
     while(!q.empty())
     {
         now = q.front() ;
         q.pop() ;
         for (i =  ; i <  ; i++)
         {
             fx = now.x + dx[i] ;
             fy = now.y + dy[i] ;
             if (fx< || fy< || fx>=  || fy>=  || map[fx][fy] == )
                 continue ;
             if (fx == ex && fy == ey)
             {
                 return now.step+ ;
             }
             t.x = fx ;
             t.y = fy ;
             t.step = now.step+ ;
             q.push(t) ;
             map[fx][fy] =  ;
         }
     }
     return  ;
 }

 int main()
 {
     //freopen("in.txt","r",stdin) ;
     while (scanf("%s %s" , &s1 , &s2) !=EOF)
     {
         sx = s1[] - 'a' ;
         sy = s1[] - '' ;
         ex = s2[] - 'a' ;
         ey = s2[] - '' ;
         printf("To get from %s to %s takes %d knight moves.\n",s1,s2,bfs()) ;
     }

     return  ;
 }