很明显的一道搜索，但是重点在怎么剪枝上。

很容易想到，如果我来过这里，并且方向是一致的，就可以不用看了。注意：一定要方向一致，毕竟可能出现交叉的情况；而且，在标记的过程中也要注意之前有没有标记。

代码

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

int T = 1;
const int N = 1500 + 10;
int n;
int ax, ay, bx, by;
char graph[N][N];
int dis[N][N];
bool vis[N][N][4];

void Bfs() {
	memset(dis, 127, sizeof(dis));
	queue<pair<int, int>> q;
	q.push({ax, ay});
	dis[ax][ay] = 0;
	while (!q.empty()) {
		int x = q.front().first, y = q.front().second;
		q.pop();
		if (!vis[x][y][0]) {
			for (int nx = x, ny = y; nx >= 1 && ny >= 1; nx--, ny--) {
				if (graph[nx][ny] == '#') break;
				if (vis[nx][ny][0]) break;
				vis[nx][ny][0] = true;
				if (dis[nx][ny] > dis[x][y] + 1) {
					dis[nx][ny] = dis[x][y] + 1;
					q.push({nx, ny});
				}
			}
		}
		if (!vis[x][y][1]) {
			for (int nx = x, ny = y; nx <= n && ny <= n; nx++, ny++) {
				if (graph[nx][ny] == '#') break;
				if (vis[nx][ny][1]) break;
				vis[nx][ny][1] = true;
				if (dis[nx][ny] > dis[x][y] + 1) {
					dis[nx][ny] = dis[x][y] + 1;
					q.push({nx, ny});
				}
			}
		}
		if (!vis[x][y][2]) {
			for (int nx = x, ny = y; nx >= 1 && ny <= n; nx--, ny++) {
				if (graph[nx][ny] == '#') break;
				if (vis[nx][ny][2]) break;
				vis[nx][ny][2] = true;
				if (dis[nx][ny] > dis[x][y] + 1) {
					dis[nx][ny] = dis[x][y] + 1;
					q.push({nx, ny});
				}
			}
		}
		if (!vis[x][y][3]) {
			for (int nx = x, ny = y; nx <= n && ny >= 1; nx++, ny--) {
				if (graph[nx][ny] == '#') break;
				if (vis[nx][ny][3]) break;
				vis[nx][ny][3] = true;
				if (dis[nx][ny] > dis[x][y] + 1) {
					dis[nx][ny] = dis[x][y] + 1;
					q.push({nx, ny});
				}
			}
		}
	}
	if (dis[bx][by] > 1e9) cout << -1;
	else cout << dis[bx][by];
}

void Solve() {
	cin >> n;
	cin >> ax >> ay >> bx >> by;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			cin >> graph[i][j];
		}
	}
	Bfs();
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	while (T--) {
		Solve();
	}
	return 0;
}