<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 {
    pair<vector<int>, vector<int>> dijkstra(int n, unordered_map<int, vector<int>> graph, unordered_map<int, vector<int>>& costs, int source)
    {
        map<int, int> Q;

        vector<int> dist(n+1, INT_MAX), prev(n+1, 0);

        dist[source] = 0; // Initialization
        Q[source] = 0; // associated priority equals 

        for (int v=1; v<n; ++v)
            if (v != source)
              Q[v] = INT_MAX;

        while (!Q.empty())
        {
            int u = Q.extract(begin(Q)).key();  // Remove and return best 

            for (size_t i=0; i< graph[u].size(); ++i)
            {
                int alt = INT_MAX;
                if (dist[u] != alt) 
                  alt = dist[u] + costs[u][i];
                int v = graph[u][i];
                if (alt < dist[v])
                {
                    prev[v] = u;
                    dist[v] = alt;
                    Q[v] = alt;
                }
            }
        }

        return { dist, prev };
    }
public:
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {

        unordered_map<int, vector<int>> graph, costs;
        for (auto& time: times)
        {
            graph[time[0]].push_back(time[1]);
            costs[time[0]].push_back(time[2]);
        }
        
        auto [dist, prev] = dijkstra(n, graph, costs, k);

        if (any_of(cbegin(dist)+1, cend(dist), [](int val) { return val == INT_MAX; }))
          return -1;

        return *max_element(cbegin(dist)+1, cend(dist));        
    }
};
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