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.
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <queue> 5 using namespace std; 6 7 8 int map[15][15]; 9 int sx, sy; 10 int ex, ey; 11 bool flag; 12 char s1[5] , s2[5] ; 13 14 int dx[] = {-2,-1,1,2,-2,-1,1,2} ; 15 int dy[] = {1,2,2,1,-1,-2,-2,-1} ; 16 17 struct node 18 { 19 int x , y , step ; 20 }; 21 22 int bfs() 23 { 24 queue<node> q ; 25 int i , fx ,fy ; 26 node now , t ; 27 now.x = sx ; 28 now.y = sy ; 29 now.step = 0 ; 30 q.push(now) ; 31 memset(map , 0 , sizeof(map)) ; 32 map[sx][sy] = 1 ; 33 while(!q.empty()) 34 { 35 now = q.front() ; 36 q.pop() ; 37 for (i = 0 ; i < 8 ; i++) 38 { 39 fx = now.x + dx[i] ; 40 fy = now.y + dy[i] ; 41 if (fx<0 || fy<0 || fx>= 8 || fy>= 8 || map[fx][fy] == 1) 42 continue ; 43 if (fx == ex && fy == ey) 44 { 45 return now.step+1 ; 46 } 47 t.x = fx ; 48 t.y = fy ; 49 t.step = now.step+1 ; 50 q.push(t) ; 51 map[fx][fy] = 1 ; 52 } 53 } 54 return 0 ; 55 } 56 57 58 59 int main() 60 { 61 //freopen("in.txt","r",stdin) ; 62 while (scanf("%s %s" , &s1 , &s2) !=EOF) 63 { 64 sx = s1[0] - 'a' ; 65 sy = s1[1] - '1' ; 66 ex = s2[0] - 'a' ; 67 ey = s2[1] - '1' ; 68 printf("To get from %s to %s takes %d knight moves.\n",s1,s2,bfs()) ; 69 } 70 71 72 73 return 0 ; 74 }
