<p>You are given a network of <code>n</code> nodes, labeled from <code>1</code> to <code>n</code>. You are also given <code>times</code>, a list of travel times as directed edges <code>times[i] = (u<sub>i</sub>, v<sub>i</sub>, w<sub>i</sub>)</code>, where <code>u<sub>i</sub></code> is the source node, <code>v<sub>i</sub></code> is the target node, and <code>w<sub>i</sub></code> is the time it takes for a signal to travel from source to target.</p>

<p>We will send a signal from a given node <code>k</code>. Return <em>the <strong>minimum</strong> time it takes for all the</em> <code>n</code> <em>nodes to receive the signal</em>. If it is impossible for all the <code>n</code> nodes to receive the signal, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/05/23/931_example_1.png" style="width: 217px; height: 239px;" />
<pre>
<strong>Input:</strong> times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
<strong>Output:</strong> 2
</pre>

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

<pre>
<strong>Input:</strong> times = [[1,2,1]], n = 2, k = 1
<strong>Output:</strong> 1
</pre>

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

<pre>
<strong>Input:</strong> times = [[1,2,1]], n = 2, k = 2
<strong>Output:</strong> -1
</pre>

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

<ul>
	<li><code>1 &lt;= k &lt;= n &lt;= 100</code></li>
	<li><code>1 &lt;= times.length &lt;= 6000</code></li>
	<li><code>times[i].length == 3</code></li>
	<li><code>1 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n</code></li>
	<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>
	<li><code>0 &lt;= w<sub>i</sub> &lt;= 100</code></li>
	<li>All the pairs <code>(u<sub>i</sub>, v<sub>i</sub>)</code> are <strong>unique</strong>. (i.e., no multiple edges.)</li>
</ul>

class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        // adj list {time, dst}
        vector<vector<vector<int>>> adj(n+1);
        // pq to track all avail path {time, dst} sorted by time
        priority_queue<vector<int>, vector<vector<int>>, decltype([](vector<int> a, vector<int> b){
            return a[0] > b[0];
        })> q;
        // time table
        vector<int> timetable(n+1,INT_MAX);

        // step1: build adjacency list, directed
        for(auto const& t: times){
            adj[t[0]].push_back({t[2], t[1]});
        }
        // step2: initiate src node
        q.push({0, k});
        timetable[k] = 0;
        int res = 0;
        // step3: 
        // - select the next nearest node and mark it visited(a node is visited when all its neighbours are added to pq)
        // - push all its neighbors, if not yet visited

        while(!q.empty()){
            auto t = q.top();
            q.pop();
            for(auto const& v: adj[t[1]]){
                int delay = t[0] + v[0];
                if(delay < timetable[v[1]]){
                    q.push({v[0]+t[0], v[1]});
                    timetable[v[1]] = delay;
                }
            }
        }
        for(int i=1; i<=n; i++){
            if(timetable[i] == INT_MAX) return -1;
            res = max(res, timetable[i]);
        }
        return res;
    }
};
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