<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:
    vector<int> distances;
    vector<pair<int,int> > adj[105];
    vector<bool> visited;
    void Dijkstra(int src,int n){
        set<pair<int,int> > s1;
        distances[src]=0;
        for(int i=1;i<=n;i++){
            s1.insert({distances[i],i});
        }
        while(s1.size()>0){
            auto itr=s1.begin();
            pair<int,int> p1=*itr;
            s1.erase(p1);
            int x=p1.first;
            int y=p1.second;
            if(x==INT_MAX){
                return;
            }
            visited[y]=true;
            for(int i=0;i<adj[y].size();i++){
                if(visited[adj[y][i].first]==false){
                    int newdist=x+adj[y][i].second;
                    if(newdist<distances[adj[y][i].first]){
                        s1.erase({distances[adj[y][i].first],adj[y][i].first});
                        distances[adj[y][i].first] = newdist;
                        s1.insert({distances[adj[y][i].first],adj[y][i].first});
                    }
                }
            }
            
        }
        return;
    }
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        for(int i=0;i<times.size();i++){
            adj[times[i][0]].push_back({times[i][1],times[i][2]});
        }
        distances.resize(n+1,INT_MAX);
        visited.resize(n+1,false);
        Dijkstra(k,n);
        int maxans=0;
        for(int i=1;i<=n;i++){
            if(distances[i]==INT_MAX){
                return -1;
            }
            maxans=max(maxans,distances[i]);
        }
        return maxans;

    }
};
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