<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>

struct Comp{
  bool operator()(vector<int>& A, vector<int>& B){
      if(A[0] != B[0])
          return A[0] >B[0];
      return true;
  }  
};



class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        vector<int>timeToSignal(n+1, INT_MAX);
        vector<vector<vector<int>>>graph(n+1);
        for(int i=0;i<times.size();i++){
            int u = times[i][0];
            int v = times[i][1];
            int w = times[i][2];
            graph[u].push_back({w, v});
        }
        priority_queue<vector<int>, vector<vector<int>>, Comp>Q;
        Q.push({0, k});
        timeToSignal[k] = 0;
        while(!Q.empty()){
            vector<int> top = Q.top();
            Q.pop();
            int w = top[0];
            int v = top[1];
            if(w > timeToSignal[v])
                continue;
            for(int i=0;i<graph[v].size();i++){
                int new_v = graph[v][i][1];
                int v_to_new_v = graph[v][i][0];
                if(timeToSignal[new_v] > w + v_to_new_v){
                    timeToSignal[new_v] = w + v_to_new_v;
                    Q.push({timeToSignal[new_v], new_v});
                }
            }
        }
        int maxTime = INT_MIN;
        for(int i=1;i<=n;i++){
            maxTime = max(maxTime, timeToSignal[i]);
        }
            
            
        return maxTime == INT_MAX ? -1 : maxTime;
    }
};
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