**最短路SPFA**
作为第一篇最短路题解，我感觉这貌似是最简单的一个思路。
我们首先把整个字符矩阵读到字符二维数组中
### 建图
* 对于$ch_{x,y}$，如果$x mod 2 == 0$且$y mod 2 == 0$，暂且管这样的一格叫做房间
* 对于每一个房间，如果它四面有一个墙壁是空格，那么向它通向的那个房间建边
### 思路
扫一遍边缘找到两个开口，以它们为起点跑SPFA，对于每个节点的结果取这两次的较小值，最终答案为所有**房间**节点的距离最大值**$+1$**
### 注意事项
本题读入毒瘤，每行需要两个`getchar()`过滤回车
### 代码
```cpp
// luogu-judger-enable-o2
// luogu-judger-enable-o2
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
const int MAXN = 205;
const int INF = 0x3f3f3f3f;
const int DX[4] = {-2, 0, 2, 0};
const int DY[4] = {0, -2, 0, 2};
const int BX[4] = {-1, 0, 1, 0};
const int BY[4] = {0, -1, 0, 1};
char mp[MAXN][MAXN];
int dis[MAXN * MAXN];
int w, h, n, m, cnt = -1;
int s[MAXN * 4];
int res[3][MAXN * MAXN];
bool vis[MAXN * MAXN];
struct Edge{
int to, val;
Edge *next;
Edge(int to, int val, Edge *next): to(to), val(val), next(next){}
};
Edge *head[MAXN * MAXN];
void AddEdge(int u, int v, int w) {
head[u] = new Edge(v, w, head[u]);
}
int Id(int x, int y) {
return m * (x - 1) + y;
}
void Spfa(int s) {
memset(vis, false, sizeof(vis));
memset(dis, INF, sizeof(dis));
dis[s] = 0;
queue<int> q; q.push(s); vis[s] = true;
while (!q.empty()) {
int u = q.front(); q.pop(); vis[u] = false;
for (Edge *e = head[u]; e; e = e->next) {
int v = e->to;
if (dis[v] > dis[u] + e->val) {
dis[v] = dis[u] + e->val;
if (!vis[v]) {
vis[v] = true;
q.push(v);
}
}
}
}
}
int main() {
cin >> w >> h;
n = h * 2 + 1; m = w * 2 + 1;
/*getchar();
for (int i = 1; i <= n; i++) {
int cnt = 0; char ch;
while (ch = getchar(), ch != '\n') mp[i][++cnt] = ch;
}*/
for (int i = 1; i <= n; i++) {
getchar();
for (int j = 1; j <= m; j++) {
mp[i][j] = getchar();
if (mp[i][j] == '\n') mp[i][j] = getchar();
}
}
for (int i = 2; i < n; i += 2) {
for (int j = 2; j < m; j += 2) {
for (int k = 0; k < 4; k++) {
int x = i + DX[k], y = j + DY[k], bx = i + BX[k], by = j + BY[k];
if (x < 1 || x > n || y < 1 || y > m || mp[bx][by] != ' ') continue;
AddEdge(Id(i, j), Id(x, y), 1);
}
}
}
for (int i = 1; i <= m; i++) {
if (mp[1][i] == ' ') s[++cnt] = Id(2, i);
if (mp[n][i] == ' ') s[++cnt] = Id(n - 1, i);
}
for (int i = 1; i <= n; i++) {
if (mp[i][1] == ' ') s[++cnt] = Id(i, 2);
if (mp[i][m] == ' ') s[++cnt] = Id(i, m - 1);
}
Spfa(s[0]);
for (int i = 1; i <= n * m; i++) res[0][i] = dis[i];
Spfa(s[1]);
for (int i = 1; i <= n * m; i++) res[1][i] = dis[i];
for (int i = 1; i <= n * m; i++) res[2][i] = min(res[1][i], res[0][i]);
int ans = 0;
for (int i = 1; i <= n * m; i++) {
if (res[2][i] < INF) ans = max(ans, res[2][i]);
}
cout << ans + 1 << endl;
return 0;
}