上课老师给了AC代码，看不懂，求解

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

int dx[] = {-2, -1, 1, 2, 2, 1, -1, -2};
int dy[] = {1, 2, 2, 1, -1, -2, -2, -1};
int n, m, sx, sy, ans[405][405];
bool vis[405][405];

struct node {
	int x, y, step;//坐标和步数 
};

bool check(int x, int y) {
	return x >= 1 && x <= n && y >= 1 && y <= m;
}

void bfs() {
	queue<node> q;//存储的是一个结构体 
	vis[sx][sy] = true;//标记走过 
	ans[sx][sy] = 0;//记录步数 
	q.push((node){sx, sy, 0});
	while (!q.empty()) {
		node tmp = q.front();
		q.pop();
		int x = tmp.x, y = tmp.y, step = tmp.step;
		for (int i = 0; i < 8; i ++) {//枚举马的8个方向 
			int nx = x + dx[i], ny = y + dy[i];
			if (check(nx, ny) && !vis[nx][ny]) {//可以访问 
				vis[nx][ny] = true;
				ans[nx][ny] = step + 1;
				q.push((node){nx, ny, step + 1});
			}
		}
	}
}

int main() {
	cin >> n >> m >> sx >> sy;
	memset(vis, false, sizeof(vis));
	for (int i = 1; i <= n; i ++) {
		for (int j = 1; j <= m; j ++) ans[i][j] = -1;
	} 
	bfs();
	for (int i = 1; i <= n; i ++) {
		for (int j = 1; j <= m; j ++) {
			printf("%-5d", ans[i][j]);
		}
		cout << endl;
	}
	return 0;
} 

~~求dalao解析~~~