<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 dijsktra(int &src, int &dest, vector<pair<int,int>>adj[], int &n){

        vector<int>distance(n+1, 1e9);
        
        // cost dest
        priority_queue< pair<int,int> , vector<pair<int,int>>, greater<pair<int,int>>> minheap;
        minheap.push({0,src});
        distance[src] = 0;

        while(!minheap.empty()){

            int cost = minheap.top().first;
            int node = minheap.top().second;
            minheap.pop();

            for(auto adj_data : adj[node]){

                int adj_node = adj_data.first;
                int adj_weight = adj_data.second;

                if(distance[adj_node] > cost + adj_weight){
                    distance[adj_node] = cost + adj_weight;
                    minheap.push({distance[adj_node],adj_node});
                }
            }
        }

        return distance[dest] == 1e9 ? -1 : distance[dest];
    }

    int networkDelayTime(vector<vector<int>>& times, int n, int k) {

        vector<pair<int,int>> adj[n+1];

        for(auto edge: times){
            int src = edge[0];
            int dest = edge[1];
            int cost = edge[2];
            
            adj[src].push_back({dest,cost});
        }

        int min_time = 0;

        for(int i=1; i<=n; i++){
            if(k != i){
                int time = dijsktra(k,i,adj,n);

                if(time == -1)
                    return -1;

                min_time = max(min_time , time );
            }
        }

        return min_time;
    }
};
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