三个小时我不知道崩溃了多少次…真的心态爆炸
这题我以为比油田简单点，结果简单搜索题并不简单…我直接裂开
有力气之后再回来补一下题解好了..

#include <iostream>
#include <queue>
#include <cstring>

#define x first
#define y second 

using namespace std;
typedef pair<int, int> PII;

const int N = 11;

int t;
int n, m;
char g[N][N];
int d[N][N];
int res ;

int dx[] = {-1, 0, 1, 0}, dy[] = {0, -1, 0, 1};
int st[N][N];

vector<PII> v;
PII p;

void dfs(int x, int y) {

    st[x][y] = 1;
    p.x = x, p.y = y;
    v.push_back(p);
    for (int i = 0; i < 4; i++) {
        int a = x + dx[i], b = y + dy[i];
        if (a < 0 || a > n - 1 || b < 0 || b > m - 1 || g[a][b] == '.' || st[a][b])
            continue;
        dfs(a, b);
    }    
    return;
}

int bfs(PII a, PII b) {
    int mx = 0, cnt = 0;
    memset(d, -1, sizeof d);
    memset(st, 0, sizeof st); 

    queue<PII> q;
    if (a != b) {
        q.push(a);
        cnt++;
    }
    q.push(b);
    cnt++;

    d[a.x][a.y] = 0, d[b.x][b.y] = 0;
    st[a.x][a.y] = 1, st[b.x][b.y] = 1;

    while (q.size()) {
        PII t = q.front();
        q.pop();

        for (int i = 0; i < 4; i++) {
            int tx = t.x + dx[i], ty = t.y + dy[i];
            if (tx < 0 || tx > n - 1 || ty < 0 || ty > m - 1 || st[tx][ty] || g[tx][ty] == '.')
                continue;
            d[tx][ty] = d[t.x][t.y] + 1;
            mx = max(mx, d[tx][ty]);
            p.x = tx, p.y = ty;
            st[tx][ty] = 1;
            q.push(p);
            cnt++;
        }
    }

        // printf("cnt: %d, mx: %d, v.size(): %d\n", cnt, mx, v.size());
        // for (int i = 0; i < n; i++) {
        //     for (int j = 0; j < m; j++) {
        //         printf("%d ", d[i][j]);
        //     }
        //     puts("");
        // }
        // puts("");
    if (cnt == v.size())
        return mx;
    return -1;

}

int main() {
    cin >> t;
    int Case = 1;
    while (t--) {
        cin >> n >> m;

        for (int i = 0; i < n; i++) 
            for (int j = 0; j < m; j++) 
                cin >> g[i][j];

        memset(st, 0, sizeof st); // 记得每轮新的测试要清空st
        memset(d, -1, sizeof d);
        v.clear();

        int f = 0; // 连通块的数量
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (g[i][j] == '#' && !st[i][j]) {
                    dfs(i, j);
                    f++;
                }
            }
        }

        res = 100; // 取个大点的值，然后找最小，肯定大不过100
        if (f > 2 || f == 0) {
            printf("Case %d: -1\n", Case++);
            continue;
        } else {

            int s = v.size();
            for (int i = 0; i < s; i++) {
                for (int j = i ; j < s; j++) {
                    int dist = bfs(v[i], v[j]);
                    if (dist != -1)
                        res = min(res, dist);
                }
            }


                // for (int i = 0; i < s; i++) {
                //     for (int j = i; j < s; j++) {
                //         bfs(v[i], v[j]);
                //         cout << "res: " << res << endl;
                //     }
                // }
            printf("Case %d: %d\n", Case++, res);
        }
    }
    return 0;
}