这里给大家提供两种解法，dfs和bfs

dfs解法

#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
char mp[N][N];
int n;
int flag = 0;
int vis[N][N];
int dx[] = {0,1,0,-1};
int dy[] = {1,0,-1,0};
void dfs(int x,int y){
    vis[x][y] = true;
    for (int i = 0; i < 4; ++i){
        int nx = x + dx[i];
        int ny = y + dy[i];
        if (nx < 1 || nx > n || ny < 1 || ny > n) continue;
        if (mp[nx][ny] == '.') continue;
        if (vis[nx][ny]) continue;
        vis[nx][ny] = 1;
        if (mp[nx - 1][ny] == '#' && mp[nx + 1][ny] == '#' && mp[nx][ny - 1] == '#' && mp[nx][ny + 1] == '#'){
            flag = 1;
        }
        dfs(nx,ny);
    }
    return;
}
int main(){
    int sum = 0;
    int cnt = 0;
    scanf("%d",&n);
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            cin >> mp[i][j];
        }
    }
    for (int i = 0; i <= n + 1; ++i){
        for (int j = 0; j <= n + 1; ++j){
            if (!mp[i][j]) mp[i][j] = '#';
        }
    }    
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            if (mp[i][j] == '#' && !vis[i][j]){
                dfs(i,j);
                sum++;
                if (flag == 1){
                    cnt++;
                    flag = 0;
                }
            }
        }
    }
    printf("%d\n",sum - cnt);
    return 0;
}

bfs解法

#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> PII;
queue<PII> q;
const int N = 1010;
int n;
int flag = 0;
char mp[N][N];
int vis[N][N];
int dx[] = {0,1,0,-1};
int dy[] = {1,0,-1,0};
void bfs(int x,int y){
    vis[x][y] = 1;
    q.push({x,y});
    while(!q.empty()){
        PII t = q.front();
        q.pop();
        for (int i = 0; i < 4; ++i) {
            int nx = t.first + dx[i];
            int ny = t.second + dy[i];
            if (nx < 1 || nx > n || ny < 1 || ny > n) continue;
            if (mp[nx][ny] == '.') continue;
            if (vis[nx][ny]) continue;
            vis[nx][ny] = 1;
            if (mp[nx - 1][ny] == '#' && mp[nx + 1][ny] == '#' && mp[nx][ny - 1] == '#' && mp[nx][ny + 1] == '#'){
                flag = 1;
            }
            q.push({nx,ny});
        }
    }
}
int main(){
    int sum = 0;
    int cnt = 0;
    scanf("%d",&n);
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j){
            cin >> mp[i][j];
        }
    }

    for (int i = 0; i <= n + 1; ++i) {
        for (int j = 0; j <= n + 1; ++j) {
          if (!mp[i][j]) mp[i][j] = '#'; 
        }
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            if (mp[i][j] == '#' && !vis[i][j]){
                bfs(i,j);
                sum ++;
                if (flag == 1){
                    cnt++;
                    flag = 0;
                }
            }
        }
    }
    printf("%d\n",sum - cnt);
    return 0;
}

所有数据均可通过