好吧我没读清楚题目，和象棋中的马还是有点不一样的，不会卡脚

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

const int N = 155;
char a[N][N], b[N][N];
int dir[8][2]  = {-1, 2, -1, -2, 1, -2, 1, 2, -2, -1, -2, 1, 2, -1, 2, 1};
int rock[8][2] = {0, 1, 0, -1, 0, -1, 0, 1, -1, 0, -1, 0, 1, 0, 1, 0};

int vis[N][N];
int n, m;

int g(int x, int y){
    return x >= 0 && x < n && y >= 0 && y < m;
}

int main(){
    scanf("%d%d", &m, &n);
    for(int i = 0; i < n; ++i){
        scanf("%s", a[i]);
    }
    int fx = -1, fy = -1;
    int hx = -1, hy = -1;
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < m; ++j){
            if(a[i][j] == 'K'){
                fx = i, fy = j;
            }
            if(a[i][j] == 'H'){
                hx = i, hy = j;
            }
        }
    }
    //printf("%d %d %d %d\n", fx, fy, hx, hy);
    queue<pair<int, int> > q;
    q.push({fx, fy});
    while(q.size()){
        int x = q.front().first;
        int y = q.front().second;
        q.pop();
        //printf("%d %d\n", x, y);
        if(x == hx && y == hy) break;
        for(int i = 0; i < 8; ++i){
            int xx = x + dir[i][0];
            int yy = y + dir[i][1];
            if(g(xx, yy) && vis[xx][yy] == 0 && a[xx][yy] != '*'){
                vis[xx][yy] = 1;
                b[xx][yy] = b[x][y] + 1;
                q.push({xx, yy});
            }
        }
    }
    printf("%d", b[hx][hy]);
    return 0;
}