<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) {
        std::vector<std::list<std::pair<int, int> > > adjList(n, std::list<std::pair<int, int> >());

        for (int i = 0; i < times.size(); ++i){
            int x = times[i][0]-1, y = times[i][1]-1, w = times[i][2];
            adjList[x].push_back(std::pair<int, int>(y, w));
        }

        std::vector<int> distance(n, std::numeric_limits<int>::max()), visited(n, 0);
        int count = n;
        distance[k-1] = 0;
        visited[k-1] = 2;

        for (int current = -1; count > 0; ){
            current = -1;
            int min_distance = std::numeric_limits<int>::max();
            for (int i = 0; i < n; ++i){
                if (visited[i] == 2 && min_distance > distance[i]){
                    current = i;
                    min_distance = distance[i];
                }
            }

            if (current == -1)
                return -1;
            
            visited[current] = 1;
            count -= 1;

            for (auto it = adjList[current].begin(); it != adjList[current].end(); ++it){
                int i = (*it).first, w = (*it).second;

                if (visited[i] != 1 && distance[current]+w < distance[i]){
                    distance[i] = distance[current]+w;
                    visited[i] = 2;
                }
            }
        }

        return *std::max_element(distance.begin(), distance.end());
    }
};
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