<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:
    const long long int INF = 1e8;

    vector<pair<int, int>> V[100000];
    long long int dis[100000];
    bool vis[100000];

    class CMP {
        public:
        bool operator()(pair<int, int> a, pair<int, int> b) {
            return (a.second > b.second);
        }
    };

    void Dijkstra(int src) {
        priority_queue<pair<int, int>, vector<pair<int, int>>, CMP> pq;
        pq.push({src, 0});
        dis[src] = 0;
        vis[src] = true;
        
        while(!pq.empty()) {
            pair<int, int> parent = pq.top(); pq.pop();
            int p_node = parent.first;
            int p_cost = parent.second;
            
            for(auto child : V[p_node]) {
                int child_node = child.first;
                int child_cost = child.second;
                
                if(p_cost + child_cost < dis[child_node]) {
                    dis[child_node] = p_cost + child_cost;
                    pq.push({child_node, dis[child_node]});
                    vis[child_node] = true;
                }
            }
        }
    }


    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        for(auto tm: times) {
            V[tm[0]].push_back({tm[1], tm[2]});
        }
        for(int i=1; i<=n; i++) {
            dis[i] = INF;
            vis[i] = false;
        }

        Dijkstra(k);

        int mx = 0;

        for(int i=1; i<=n; i++) {
            if(dis[i] == INF) {
                return -1;
            }
            if(mx < dis[i]) {
                mx = dis[i];
            }
        }

        return mx;
    }
};
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