润色这段文字：If the transmission line is surrounded by uniform medium, the capacitance matrix per unit length can be obtained from the inductance matrix

per unit length using the reciprocal relation.

我怎么可以知道什么是“润色”？但是我可以帮你修改这段文字：If the transmission line is surrounded by a uniform medium, you can obtain the capacitance matrix per unit length from the inductance matrix per unit length using the reciprocal relation.

相关问题

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!

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