from collections import deque
# 获取输入信息
C, R = map(int, input().split())

# 绘制地图
graph = []
for _ in range(R):
    row = input()
    graph.append(row)

for r in range(R):
    for c in range(C):
        if graph[r][c] == 'K':
            start_x, start_y = r, c

# 偏移量
dx = [-2, -1, 1, 2, 2, 1, -1, -2]
dy = [1, 2, 2, 1, -1, -2, -2, -1]


def bfs():
    queue = deque([(start_x, start_y, 0)])
    visited = [[False] * C for _ in range(R)]
    visited[start_x][start_y] = True
    while queue:
        x, y, step = queue.popleft()
        if graph[x][y] == 'H':
            return step
        for i in range(8):
            new_x = x + dx[i]
            new_y = y + dy[i]
            if 0 <= new_x < R and 0 <= new_y < C and not visited[new_x][new_y] and graph[new_x][new_y]!= '*':
                visited[new_x][new_y] = True
                queue.append((new_x, new_y, step + 1))

print(bfs())

首先读取农场的列数和行数以及地图信息，找到马的起始位置。然后采用BFS算法，初始化队列和访问标记列表，从马的位置开始探索八个方向，当找到牛的位置时返回最短路径步数。通过不断从队列中取出位置进行探索，直到找到目标或队列为空。

一道走迷宫的题目，位移偏移量用马走日的方式进行。然后因为地图大小150*150，因此选择bfs走迷宫。

#include<bits/stdc++.h>
using namespace std;

const int N = 155;
const int dx[] = {-2, -1, 1, 2, 2, 1, -1, -2};
const int dy[] = {-1, -2, -2, -1, 1, 2, 2, 1};

int n, m;
char g[N][N];
int mark[N][N];
int sx, sy, tx, ty;

void bfs(){
    queue<pair<int, int>> q;
    q.push({sx, sy});
    mark[sx][sy] = 1;
    
    while(q.size()){
        int x = q.front().first;
        int y = q.front().second;
        q.pop();
        
        for(int i=0;i<8;i++){
            int a = x + dx[i];
            int b = y + dy[i];
            if(a < 0 || b < 0) continue;
            if(a >= n || b >= m) continue;
            if(g[a][b] == '*' || mark[a][b]) continue;
            
            q.push({a, b});
            mark[a][b] = mark[x][y] + 1;
            if(a == tx && b == ty) return ;
        }
    }
}

int main(){
    cin>>m>>n;
    for(int i=0;i<n;i++) cin>>g[i];
    
    for(int i=0;i<n;i++)for(int j=0;j<m;j++)
    if(g[i][j] == 'K') {sx = i; sy = j;}
    else if(g[i][j] == 'H') {tx = i; ty = j;}
    
    bfs();
    cout<<mark[tx][ty] - 1<<endl;
    return 0;
}