<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>

#include<queue>
#include <iostream> 
#include <algorithm> 

class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {

        vector<vector<int>> timesCor;

        for(vector<int> vec: times)
        {
            timesCor.push_back({vec[0] - 1, vec[1] - 1, vec[2]});
        }

        vector<vector<int>> edgesNTimes(n,vector<int>(n, -1));

        for(vector<int> vec:timesCor)
        {
            edgesNTimes[vec[0]][vec[1]] = vec[2];
        }    

        vector<int> finalTimes(n, INT_MAX);
        finalTimes[k - 1] = 0;

        vector<bool> visited(n , false);

        queue<int> q;
        q.push(k - 1);

        while(!q.empty())
        {
            int front = q.front();
            q.pop();

        for(int i = 0; i < edgesNTimes[front].size(); i++)
        {
            if(edgesNTimes[front][i] == -1)
                continue;
            
            int curTime = finalTimes[front]+ edgesNTimes[front][i];

            if(curTime < finalTimes[i])
            {
                q.push(i);
                finalTimes[i] = curTime;
            }
        }

        }

        return *std::max_element(finalTimes.begin(), finalTimes.end())==INT_MAX ? -1 : *std::max_element(finalTimes.begin(), finalTimes.end());
    }
};
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