<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 dijkstra(vector<vector<vector<int>>> &graph, int &n, int s){
        typedef pair<int,int> pr;
        priority_queue<pr, vector<pr>, greater<pr>> p;
        vector<int> dist(n, INT_MAX);

        dist[s]=0;
        p.push({0,s});

        while(p.size()){
            int weight=p.top().first;
            int node=p.top().second;

            p.pop();

            for(auto &adjNodes: graph[node]){
                int edgeWeight=adjNodes[1];
                int adjNode=adjNodes[0];

                if(edgeWeight+weight < dist[adjNode]){
                    dist[adjNode]=edgeWeight+weight;
                    p.push({dist[adjNode],adjNode});
                }
            }
        }

        int result=INT_MIN;
        for(int &d: dist){
            if(d==INT_MAX) return -1;
            result=max(result,d);
        }
        return result;
    }


    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        
      
        vector<vector<vector<int>>> graph(n);
        for(auto vec: times){
            graph[vec[0]-1].push_back({vec[1]-1,vec[2]});
        }

        return dijkstra(graph, n,k-1);
    


    }
};
56265