<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) {
        int maxDistance = 0;
        vector<int> distances(n + 1, INT_MIN);
        distances[k] = 0;
        
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> heap;
        vector<vector<vector<int>>> network(n + 1);
        
        for (vector<int> edge: times){
            if (edge[0] == k){
                heap.push({edge[2], edge[1]});
                continue;
            } 
            network[edge[0]].push_back({edge[1], edge[2]});
        }
        
        while (!heap.empty()){            
            vector<int> cur = heap.top();
            heap.pop();
                                    
            distances[cur[1]] = cur[0];
            for (vector<int> edge: network[cur[1]]){
                if (distances[edge[0]] == INT_MIN) heap.push({distances[cur[1]] + edge[1], edge[0]});
            } 
            
            while (heap.size() > 0 && distances[heap.top()[1]] != INT_MIN) heap.pop();
        }
        
       for (int i = 1; i <= n; i++){
           if (distances[i] == INT_MIN) return -1;
           maxDistance = max(maxDistance, distances[i]);
       }
        
        return maxDistance;
    }
};
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