<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:
struct cmp{
    bool operator()(pair<int,int> a, pair<int,int> b){
        return a.first>=b.first; // need to keep all max elements inside the Heap so we can pop Min element
    }
};
int networkDelayTime(vector<vector<int>>& times, int n, int k) 
{
    unordered_map<int,unordered_map<int,int>> mp;
    for(auto time : times)
        mp[time[0]][time[1]]=time[2];
    
    priority_queue<pair<int,int>,vector<pair<int,int>>,cmp> pq;
    pq.push({0,k});
    
    vector<bool> visited(n+1);
    int res = 0;
    
    while(!pq.empty())
    {
        pair<int,int> cur = pq.top();   pq.pop();
        int curDist = cur.first, curNode = cur.second;
        
        if(visited[curNode]) continue;
        visited[curNode] = true;

        n--;
        res = curDist;

        if(mp.count(curNode))
        {
            for(auto mp2: mp[curNode])
                pq.push( {curDist+mp2.second , mp2.first} );
        }
    }
    return n==0 ? res : -1;
}
};
52953