<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) {
        unordered_map<int, vector<pair<int, int>>> edges;
        for (const auto &time :times) {
            edges[time[0] - 1].push_back(make_pair(time[1] - 1, time[2]));
        }
        priority_queue<
            pair<int, int>, 
            vector<pair<int, int>>, 
            greater<pair<int, int>>> pq;
        pq.push(make_pair(0, k - 1));
        unordered_set<int> visited;
        visited.insert(k - 1);
        int max_total_weight = 0;
        while (visited.size() != n && !pq.empty()) {
            pair<int, int> nearest = pq.top();
            pq.pop();
            int total_weight = nearest.first;
            int nearest_node = nearest.second;
            visited.insert(nearest_node);
            max_total_weight = max(total_weight, max_total_weight);
            for (const auto &edge : edges[nearest_node]) {
                int next_node = edge.first;
                int edge_weight = edge.second;
                if (visited.count(next_node) > 0) {
                    continue;
                }
                pq.push(make_pair(total_weight + edge_weight, next_node));
            }
        }
        return visited.size() != n ? -1 : max_total_weight;
    }
};
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