You are given a map of a building made up of n×m squares, and each square is either a floor or a wall. Find how many rooms are in the building. A room is defined by a group of any number of adjacently connected floor squares, that is surrounded by walls. Input The first input line has two integers n and m: the height and width of the map. Then there are n lines each containing m characters describing the map. Each character is either . (floor) or # (wall). Output Print one integer, the number of rooms. Constraints 1≤n,m≤1000 Sample Input: 5 8 ######## #..#...# ####.#.# #..#...# ######## Output: 3

Sure! Here's a C++ code implementation that solves the problem:

cpp
#include <iostream>
#include <vector>

using namespace std;

int n, m;
vector<vector<char>> building;
vector<vector<bool>> visited;

// Define possible movements: up, down, left, right
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};

// Function to perform DFS to find rooms
void explore(int x, int y) {
visited[x][y] = true;

// Explore all possible neighboring cells
for (int i = 0; i < 4; i++) {
int nx = x + dx[i];
int ny = y + dy[i];

// Check if neighboring cell is within bounds and unvisited
if (nx >= 0 && nx < n && ny >= 0 && ny < m && building[nx][ny] == '.' && !visited[nx][ny]) {
explore(nx, ny);
}
}
}

int countRooms() {
int roomCount = 0;

// Initialize visited matrix
visited.assign(n, vector<bool>(m, false));

// Traverse through the building and count rooms
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (building[i][j] == '.' && !visited[i][j]) {
roomCount++;
explore(i, j);
}
}
}

return roomCount;
}

int main() {
cin >> n >> m;

// Read the building map
building.assign(n, vector<char>(m));
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> building[i][j];
}
}

int roomCount = countRooms();
cout << roomCount << endl;

return 0;
}

You can compile and run this code to get the desired output. Let me know if you have any further questions!