Editing 2016 Dec Silver Problem 3 Moocast (section)

== C++ Implementation ==

First, let's create a function to check if cow A can transmit a message to cow B:

<pre>
bool canTransmit(const vector<int>& cowA, const vector<int>& cowB) {
    int dx = cowA[0] - cowB[0];
    int dy = cowA[1] - cowB[1];
    int distanceSquared = dx * dx + dy * dy;
    return distanceSquared <= cowA[2] * cowA[2];
}

</pre>

Next, implement the BFS traversal to find the number of reachable cows:
<pre>
int bfs(const vector<vector<int>>& cows, int start) {
    int count = 0;
    vector<bool> visited(cows.size(), false);
    queue<int> q;
    q.push(start);
    visited[start] = true;

    while (!q.empty()) {
        int current = q.front();
        q.pop();
        count++;

        for (int i = 0; i < cows.size(); ++i) {
            if (!visited[i] && canTransmit(cows[current], cows[i])) {
                visited[i] = true;
                q.push(i);
            }
        }
    }
    return count;
}

</pre>

Finally, iterate over all cows and find the maximum number of reachable cows:

<pre>
int main() {
    int n;
    cin >> n;
    vector<vector<int>> cows(n, vector<int>(3));
    for (int i = 0; i < n; ++i) {
        cin >> cows[i][0] >> cows[i][1] >> cows[i][2];
    }

    int maxCows = 0;
    for (int i = 0; i < n; ++i) {
        maxCows = max(maxCows, bfs(cows, i));
    }

    cout << maxCows << endl;
}

</pre>