#include <bits/stdc++.h>
using namespace std;
int m, n;
int a[1001][1001];//记录地图
bool b[1001][1001];//此点是否被访问
int c[1001][1001];//通过地图的坐标，记录公主到灌木丛，再到骑士的距离之和
int x, y;//公主坐标
int x1, y11;//骑士坐标
int sum,sum1;//记录灌木丛的长度用
struct point
{
	int x, y;
	int length;
};
int p1 = 0;
int dx[] = { 0,0,1,-1 };
int dy[] = { 1,-1,0,0 };
int main()
{
	cin >> m >> n;
	for (int i = 1;i <= n;i++)
		for (int j = 1;j <= m;j++)
		{
			cin >> a[i][j];
			if (a[i][j] == 2)x = j, y = i;
			if (a[i][j] == 3)x1 = j, y11 = i;
			if (a[i][j] == 4)sum++;
		}
	queue<point> s1;
	queue<point> s2;
	s1.push({ x,y,0});
	while (s1.size())//从公主开始做BFS，直至获取到达所以灌木丛的最短距离
	{
		
		int x2 = s1.front().x,
			y2 = s1.front().y,
			length2=s1.front().length;
		b[y2][x2] = 1;
		if (a[y2][x2]==4)
		{
			s2.push(s1.front());
			c[y2][x2] = s1.front().length;
			sum1++;
			if (sum == sum1)break;//获得全部的灌木路径，就退出
		}
		for (int i = 0;i < 4;i++)
		{
			if (x2 + dx[i] >= 1 && x2 + dx[i] <= m && y2 + dy[i] >= 1 && y2 + dy[i] <= n && !b[y2 + dy[i]][x2 + dx[i]]&&a[y2 + dy[i]][x2 + dx[i]]!=1&& a[y2 + dy[i]][x2 + dx[i]] != 3)
				s1.push({ x2 + dx[i] ,y2 + dy[i] ,length2 + 1 });
		}
		s1.pop();
	}
	sum1 = 0;
	queue<point> s3;//从骑士出发，找到各个灌木丛距离骑士的距离。将这些长度与之前求得的公主到每个灌木丛的距离相加，即可得出所有灌木丛条件下的长度
	s3.push({ x1,y11,0 });
	memset(b, 0, sizeof(b));
	while (s3.size())
	{
		int x2 = s3.front().x,
			y2 = s3.front().y,
			length2 = s3.front().length;
		b[y2][x2] = 1;
		if (a[y2][x2] == 4)
		{
			c[y2][x2]+= s3.front().length;
			sum1++;
			if (sum == sum1)break;//获得全部的灌木路径，就退出
		}
		for (int i = 0;i < 4;i++)
		{
			if (x2 + dx[i] >= 1 && x2 + dx[i] <= m && y2 + dy[i] >= 1 && y2 + dy[i] <= n && !b[y2 + dy[i]][x2 + dx[i]] && a[y2 + dy[i]][x2 + dx[i]] != 1)
				s3.push({ x2 + dx[i] ,y2 + dy[i] ,length2 + 1 });
		}
		s3.pop();
	}
	int min1 = 0x7fffffff;
	for (int i = 1;i <= sum;i++)//比较长度
	{
		min1 = min(min1, c[s2.front().y][s2.front().x]);
		s2.pop();
	}
	cout << min1;
}

只对了2个其它的不是爆内存就是错误，请大佬指正