<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 Solution {
public:
    typedef pair<int, int> ii;
    vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) {
        multiset<ii> ms;
        ms.insert({0, 0});
        ms.insert({0, firstPerson});
        vector<int> know(n, 1e9);
        know[0] = 0, know[firstPerson] = 0;
        vector<vector<ii>> edges(n);
        for (auto &v: meetings) {
            int x = v[0], y = v[1], t = v[2];
            edges[x].push_back({y, t});
            edges[y].push_back({x, t});
        }
        for (int i = 1; i < n; i++) {
            if (i == firstPerson) continue;
            ms.insert({1e9, i});
        }
        vector<int> ans;
        vector<int> done(n, 0);
        while (!ms.empty()) {
            auto [t, p] = *(ms.begin());
            if (t == 1e9) break;
            ans.push_back(p);
            done[p] = 1;
            ms.erase(ms.begin());
            for (auto [v, tmp]: edges[p]) {
                if (tmp < t) continue;
                if (done[v]) continue;
                ms.erase({know[v], v});
                know[v] = min(know[v], tmp);
                ms.insert({know[v], v});
            }
        }
        return ans;
    }
};

// class Solution {
// public:
//     typedef pair<int, int> ii;
//     vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) {
//         set<ii> st;
//         map<int, int> mp;
//         st.insert({0, 0});
//         st.insert({0, firstPerson});
//         mp[0] = 0;
//         mp[firstPerson] = 0;
//         vector<int> ans;
//         vector<vector<ii>> edges(n, vector<ii> (0, {-1, 1}));
//         for (int i = 0; i < meetings.size(); i++) {
//             edges[meetings[i][0]].push_back({meetings[i][1], meetings[i][2]});
//             edges[meetings[i][1]].push_back({meetings[i][0], meetings[i][2]});
//         }
//         int time = 0;
//         while (!st.empty()) {
//             auto it = st.begin();
//             ans.push_back(it->second);
//             int x = it->second;
//             time = it->first;
//             st.erase(*it);
//             for (auto edge: edges[x]) {
//                 if (edge.second < time) {
//                     continue;
//                 }
//                 if (mp.count(edge.first) != 0) {
//                     if (mp[edge.first] <= edge.second) {
//                         continue;
//                     }
//                 }
//                 st.erase({mp[edge.first], edge.first});
//                 mp[edge.first] = edge.second;
//                 st.insert({edge.second, edge.first});
//             }
//         }
//         return ans;
//     }
// };
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