奶牛们去一个 N × M N\times M N×M 玉米迷宫， 2 ≤ N ≤ 300 , 2 ≤ M ≤ 300 2 \leq N \leq 300,2 \leq M \leq300 2≤N≤300,2≤M≤300。
迷宫里有一些传送装置，可以将奶牛从一点到另一点进行瞬间转移。这些装置可以双向使用。
如果一头奶牛处在这个装置的起点或者终点，这头奶牛就必须使用这个装置。
玉米迷宫除了唯一的一个出口都被玉米包围。
迷宫中的每个元素都由以下项目中的一项组成：
# 表示，这些格子是不可以通过的。. 表示，可以简单的通过。= 表示。@ 表示奶牛能在一格草地上可能存在的四个相邻的格子移动，花费 1 1 1 个单位时间。从装置的一个结点到另一个结点不花时间。
This past fall, Farmer John took the cows to visit a corn maze. But this wasn’t just any corn maze: it featured several gravity-powered teleporter slides, which cause cows to teleport instantly from one point in the maze to another. The slides work in both directions: a cow can slide from the slide’s start to the end instantly, or from the end to the start. If a cow steps on a space that hosts either end of a slide, she must use the slide.
The outside of the corn maze is entirely corn except for a single exit.
The maze can be represented by an N x M (2 <= N <= 300; 2 <= M <= 300) grid. Each grid element contains one of these items:
* Corn (corn grid elements are impassable)
* Grass (easy to pass through!)
* A slide endpoint (which will transport a cow to the other endpoint)
* The exit
A cow can only move from one space to the next if they are adjacent and neither contains corn. Each grassy space has four potential neighbors to which a cow can travel. It takes 1 unit of time to move from a grassy space to an adjacent space; it takes 0 units of time to move from one slide endpoint to the other.
Corn-filled spaces are denoted with an octothorpe (#). Grassy spaces are denoted with a period (.). Pairs of slide endpoints are denoted with the same uppercase letter (A-Z), and no two different slides have endpoints denoted with the same letter. The exit is denoted with the equals sign (=).
Bessie got lost. She knows where she is on the grid, and marked her current grassy space with the ‘at’ symbol (@). What is the minimum time she needs to move to the exit space?
第一行：两个用空格隔开的整数 N N N 和 M M M。
第 2 ∼ N + 1 2\sim N+1 2∼N+1 行：第 i + 1 i+1 i+1 行描述了迷宫中的第 i i i 行的情况（共有 M M M个字符，每个字符中间没有空格）。
一个整数，表示起点到出口所需的最短时间。
5 6
###=##
#.W.##
#.####
#.@W##
######
3
例如以下矩阵， N = 5 , M = 6 N=5,M=6 N=5,M=6。
###=##
#.W.##
#.####
#.@W##
######
唯一的一个装置的结点用大写字母 W \tt{W} W 表示。
最优方案为：先向右走到装置的结点，花费一个单位时间，再到装置的另一个结点上，花费 0 0 0 个单位时间，然后再向右走一个，再向上走一个，到达出口处，总共花费了 3 3 3 个单位时间。
#include<bits/stdc++.h>
using namespace std;
struct node {
int x, y, step;
node(int xx, int yy, int ste) {
x = xx, y = yy, step = ste;
}
};
int n, m;
char a[305][305];
int vis[305][305];
queue<node> q;
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
pair<int, int> findDoor(char c, int x, int y) {
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
//不能找到它本身
if (a[i][j] == c && (i != x || j != y))
return {i, j};
}
}
}
void bfs() {
while (!q.empty()) {
node now = q.front();
q.pop();
int x = now.x, y = now.y, step = now.step;
if (a[x][y] == '=') {
cout << step;
return;
}
for (int i = 0; i < 4; i++) {
int xx = x + dx[i];
int yy = y + dy[i];
if (xx >= 0 && yy >= 0 && xx < n && yy < m && !vis[xx][yy] && a[xx][yy] != '#') {
vis[xx][yy] = 1;
//传送门, 把传送到的点放入队列（传送门自己本身可以不加队列里，把它到的目的传送门加队列）
if (a[xx][yy] >= 'A' && a[xx][yy] <= 'Z') {
//获取传送的目的地
pair<int, int> p = findDoor(a[xx][yy], xx, yy);
xx = p.first, yy = p.second;
//传送门不标记，随便传
vis[p.first][p.second] = 0;
}
//(xx,yy)可能是另一半的传送门，也可能是草地'.'
q.push(node(xx, yy, step + 1));
}
}
}
}
int main() {
cin >> n >> m;
int startx, starty;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
cin >> a[i][j];
if (a[i][j] == '@')
startx = i, starty = j;
}
}
q.push(node(startx, starty, 0));
vis[startx][starty] = 1;
bfs();
return 0;
}