<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 Compair {
public:
    bool operator()(vector<int>& a, vector<int>& b) {
        return a[0] > b[0];
    }
};

class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        // Since we will use priority queue anyways, the children receiving times
        // don't need to be sorted.
        unordered_map<int, unordered_map<int, vector<int>>> m;
        for (auto& v : times) {
            m[v[0]][v[2]].push_back(v[1]);
        }
        vector<bool> visited(n + 1, false);
        // Queue of {time, node}
        priority_queue<vector<int>, vector<vector<int>>, Compair> q;
        q.push(vector<int>{0, k});
        int max_time = 0;
        while (q.size() > 0) {
            vector<int> cur_pair = q.top();
            q.pop();
            int cur_time = cur_pair[0];
            int cur_node = cur_pair[1];
            if (visited[cur_node]) continue;
            visited[cur_node] = true;
            max_time = max(max_time, cur_time);
            // This for loop doesn't need to be in order.
            for (auto& pair : m[cur_node]) {
                for (int next_node : pair.second) {
                    // cout << "pushing " << next_node << endl;
                    q.push(vector<int>{pair.first + cur_time, next_node});
                }
            }
        }

        for (int i = 1; i <= n; i++) {
            if (!visited[i]) {
                return -1;
            }
        }
        return max_time;
    }
};
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