poj 2243 Knight Moves

这题和poj 1915一样，用bfs做走马步。现在再看当时的代码，真是好幼稚啊。
 1 #include <stdio.h>
 2 #include <string.h>
 3 int X[] = {-2,-2,-1,-1,1,1,2,2},
 4     Y[] = {-1,1,-2,2,-2,2,-1,1};
 5 struct Point
 6 {
 7     int x,y;
 8     void init(int a,int b)
 9     {x = a; y = b;}
10 }q[70],s,t;
11 int head,tail,map[9][9];
13 void bfs()
14 {
15     Point u,v;
16     int x1,y1,i;
17     head = tail = 0;
18     q[0] = s;
19     map[s.x][s.y] = 0;
20     if(s.x == t.x && s.y == t.y)
21     {
22         printf("To get from %c%d to %c%d takes 0 knight moves.\n"
23             ,s.x-1+'a',s.y,t.x-1+'a',t.y);
24         return ;
25     }
26     while(head <= tail)
27     {
28         u = q[head++];
29         for(i = 0; i < 8; i++)
30         {
31             x1 = u.x + X[i];
32             y1 = u.y + Y[i];
33             if(x1 > 0 && x1 < 9 && y1 > 0 && y1 < 9 && map[x1][y1] == -1)
34             {
35                 if(x1 == t.x && y1 == t.y)
36                 {    printf("To get from %c%d to %c%d takes %d knight moves.\n"
37                     ,s.x-1+'a',s.y,t.x-1+'a',t.y,map[u.x][u.y]+1);return ;}
38                 v.init(x1,y1);
39                 q[++tail] = v;
40                 map[x1][y1] = map[u.x][u.y] + 1;
41             }
42         }
43     }
44 }
45 int main()
46 {
47     char t1,t2;
48     while(~scanf("%c%d %c%d",&t1,&s.y,&t2,&t.y))
49     {
50         getchar();
51         s.x = t1 -'a'+1;
52         t.x = t2 -'a'+1;
53         memset(map,-1,sizeof map);
54         bfs();
55     }
56     return 0;
57 }