<p>You are given an integer <code>n</code> indicating there are <code>n</code> people numbered from <code>0</code> to <code>n - 1</code>. You are also given a <strong>0-indexed</strong> 2D integer array <code>meetings</code> where <code>meetings[i] = [x<sub>i</sub>, y<sub>i</sub>, time<sub>i</sub>]</code> indicates that person <code>x<sub>i</sub></code> and person <code>y<sub>i</sub></code> have a meeting at <code>time<sub>i</sub></code>. A person may attend <strong>multiple meetings</strong> at the same time. Finally, you are given an integer <code>firstPerson</code>.</p>

<p>Person <code>0</code> has a <strong>secret</strong> and initially shares the secret with a person <code>firstPerson</code> at time <code>0</code>. This secret is then shared every time a meeting takes place with a person that has the secret. More formally, for every meeting, if a person <code>x<sub>i</sub></code> has the secret at <code>time<sub>i</sub></code>, then they will share the secret with person <code>y<sub>i</sub></code>, and vice versa.</p>

<p>The secrets are shared <strong>instantaneously</strong>. That is, a person may receive the secret and share it with people in other meetings within the same time frame.</p>

<p>Return <em>a list of all the people that have the secret after all the meetings have taken place. </em>You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 6, meetings = [[1,2,5],[2,3,8],[1,5,10]], firstPerson = 1
<strong>Output:</strong> [0,1,2,3,5]
<strong>Explanation:
</strong>At time 0, person 0 shares the secret with person 1.
At time 5, person 1 shares the secret with person 2.
At time 8, person 2 shares the secret with person 3.
At time 10, person 1 shares the secret with person 5.​​​​
Thus, people 0, 1, 2, 3, and 5 know the secret after all the meetings.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 4, meetings = [[3,1,3],[1,2,2],[0,3,3]], firstPerson = 3
<strong>Output:</strong> [0,1,3]
<strong>Explanation:</strong>
At time 0, person 0 shares the secret with person 3.
At time 2, neither person 1 nor person 2 know the secret.
At time 3, person 3 shares the secret with person 0 and person 1.
Thus, people 0, 1, and 3 know the secret after all the meetings.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 5, meetings = [[3,4,2],[1,2,1],[2,3,1]], firstPerson = 1
<strong>Output:</strong> [0,1,2,3,4]
<strong>Explanation:</strong>
At time 0, person 0 shares the secret with person 1.
At time 1, person 1 shares the secret with person 2, and person 2 shares the secret with person 3.
Note that person 2 can share the secret at the same time as receiving it.
At time 2, person 3 shares the secret with person 4.
Thus, people 0, 1, 2, 3, and 4 know the secret after all the meetings.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= meetings.length &lt;= 10<sup>5</sup></code></li>
	<li><code>meetings[i].length == 3</code></li>
	<li><code>0 &lt;= x<sub>i</sub>, y<sub>i </sub>&lt;= n - 1</code></li>
	<li><code>x<sub>i</sub> != y<sub>i</sub></code></li>
	<li><code>1 &lt;= time<sub>i</sub> &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= firstPerson &lt;= n - 1</code></li>
</ul>

class DJ {
public:
    DJ(int N) : N(N) {
        init();
    }
    vector<int> parent, rank, count, point;

    void init() {
        parent.resize(N);
        rank.resize(N);
        count.resize(N);
        point.resize(N);
        for (int i=0; i<N; i++) {
            parent[i] = i;
            rank[i] = 0;
            count[i] = 1;
            point[i] = 0;
        }
    }

    int find(int u) {
        if (parent[u] != u) parent[u] = find(parent[u]);
        return parent[u];
    }

    void join(int u, int v) {
        u = find(u);
        v = find(v);
        if (u == v) return;
        if (rank[u] == rank[v]) rank[u]++;
        if (rank[u] < rank[v]) {
            parent[u] = v;
            count[v] += count[u];
        }
        else {
            parent[v] = u;
            count[u] += count[v];
        }
    }

    int getCount(int i) {
        int parent = find(i);
        return count[parent];
    }

    int getPoint(int i) {
        int parent = find(i);
        return point[parent];
    }
    void addPoint(int i, int val) {
        int parent = find(i);
        point[parent] += val;
    }
    void reset(int i) {
        parent[i] = i;
        rank[i] = 0;
        count[i] = 1;
        point[i] = 0;
    }
private:
    int N;
};

class Solution {
public:
    vector<int> findAllPeople(int n, vector<vector<int>>& a, int firstPerson) {
        // sort(a.begin(), a.end(), [](vector<int>& a1, vector<int>& a2) {
        //     return a1[2] < a2[2];
        // });
        map<int, vector<vector<int>>> m;
        for (auto e : a) m[e[2]].push_back({e[0], e[1]});

        DJ dj(n + 1);
        dj.join(0, firstPerson);
        for (auto ee : m) {
            for (auto e : ee.second) {
                int u = e[0];
                int v = e[1];
                dj.join(u, v);
            }
            for (auto e : ee.second) {
                int u = e[0];
                int v = e[1];
                if (dj.find(u) != dj.find(0)) {
                    dj.reset(u);
                    dj.reset(v);
                }
            }
        }
        vector<int> ans;
        for (int i = 0; i < n; i++) {
            if (dj.find(i) == dj.find(0)) ans.push_back(i);
        }
        return ans;
    }
};
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