CSES :: Graph Algorithms :: Counting Rooms

Problem : Please find the problem here.

Explanation : considering the given structure as graph, with each '.' character being an edge between ith and jth node, the number of rooms is equal to number of connected components in the graph.

Code : Used DFS to count number of connected components.

#include <bits/stdc++.h>

using namespace std;

const int mxN = 1e3;
const int di[4] = {1, 0, -1, 0}, dj[4] = {0, 1, 0, -1};
int n, m;
string adj[mxN];

bool isValid(int i, int j) {
    return (i >= 0 && j >= 0 && i < n && j < m && adj[i][j]=='.');
}

void DFS(int i, int j) {
    adj[i][j] = '#';

    for (int k = 0; k < 4; k++) {
        if (isValid(i + di[k], j + dj[k])) {
            DFS(i + di[k], j + dj[k]);
        }

    }
}

int countRooms() {
    int rooms = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (adj[i][j] == '.') {
                DFS(i, j);
                rooms++;
            }
        }
    }

    return rooms;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin >> n >> m;

    for (int i = 0; i < n; i++) {
        cin >> adj[i];
    }

    cout << countRooms() << '\n';

    return 0;
}

CSES :: Introductory Problems :: Weird Algorithm

Problem : Please find the problem here.

Explanation : 

Code : 

#include <bits/stdc++.h>
 
using namespace std;
 
void solve() {
    int n;
    cin >> n;
 
    cout << n << ' ';
    while (n != 1)
    {
        n = n & 1 ? n / 2 : n * 3 + 1;
        cout << n << ' ';
    }
}
 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
 
    solve();
 
    return 0;
}