58分BFS求救，有详细注释

#include <iostream>
#include <string>
#include <queue>
#include <algorithm>
using namespace std;
const int N = 505 , M = 26;
typedef pair<int,int> pll;
int n , m  , a[M] , ans[N][N];
char s[N][N];
pll point1[M] , point2[M] , S , T;
bool st[N][N];
bool check(int x,int y){
	if(x < 1 || x > n || y < 1 || y > m) return false;
	if(s[x][y] == '#' || st[x][y]) return false;
	return true;
}
void bfs(int i, int j)
{
	int dx[] = {1 , -1 , 0 , 0};
	int dy[] = {0 , 0 , 1 , -1};
	queue<pll> q;
	q.push({i , j});
	st[i][j] = true;
	while(q.size())
	{
		auto t = q.front();
		q.pop();
		//到达终点
		if(s[t.first][t.second] == '=')
		{
			cout << ans[t.first][t.second];
			return ;
		}
		if(s[t.first][t.second] >= 'A' && s[t.first][t.second] <= 'Z'){
			//取出到这个点的时间
			int k = ans[t.first][t.second];
			//取出俩个传送门的地址
			auto t1 = point1[s[t.first][t.second] - 'A'];
			auto t2 = point2[s[t.first][t.second] - 'A'];
			//判断当前在哪个传送门
			if(t == t1) t = t2;
			else t = t1;
			st[t.first][t.second] = true;
			//传送过去后耗时应该相同
			ans[t.first][t.second] = k;
		}
		//遍历上下左右四个方向
		for (int i = 0 ; i <= 3 ; ++i)
		{
			int rx = t.first + dx[i] , ry = t.second + dy[i];
			//如果坐标合法
			if(check(rx,ry)){
				//入队
				q.push({rx , ry});
				//记为已到达
				st[rx][ry] = true;
				//
				ans[rx][ry] = ans[t.first][t.second] + 1;				
			}	
		}
	}
}
int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0),cout.tie(0);
	cin >> n >> m;
	for (int i = 1 ; i <= n ; ++i)
	{
		for (int j = 1 ; j <= m ; ++j){
			cin >> s[i][j];
			if(s[i][j] == '=')
			{
				T = {i , j};
			}
			if(s[i][j] == '@')
			{
				S = {i , j};
			}
        //记录传送点
			if(s[i][j] >= 'A' && s[i][j] <= 'Z')
			{
				int t = s[i][j] - 'A';
				if(a[t] == 0) {
					a[t] = 1;
					point1[t] = {i , j};
				}
				else{
					point2[t] = {i , j};
				}
			}
		}		
	}
	bfs(S.first , S.second);
	return 0;
}