<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>
class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        
        // Get a lookup to map node to starting edges
        vector<vector<int>> connections_from_node(n);
        for(int i = 0; i < times.size(); i++){
            connections_from_node[times[i][0] - 1].push_back(i);
        }
        
        // Apply graph traversal
        vector<int> time_to_reach(n,INT_MAX);
        time_to_reach[k-1] = 0;
        
        queue<int> nodes_to_visit;
        nodes_to_visit.push(k-1);
        
        while (!nodes_to_visit.empty()){
            int current_node = nodes_to_visit.front();
            nodes_to_visit.pop();
            int cost = time_to_reach[current_node];
            
            for(auto connection: connections_from_node[current_node]){
                int destination = times[connection][1] - 1;
                int weight = times[connection][2];
                
                if(weight + cost < time_to_reach[destination]){
                    time_to_reach[destination] = weight + cost;
                    nodes_to_visit.push(destination);
                }
            }
        }
        
        int max_time_to_reach_all = *max_element(time_to_reach.begin(), time_to_reach.end());
        
        if(max_time_to_reach_all == INT_MAX){
            return -1;
        } else {
            // int sum = 0;
            // for(auto t: time_to_reach) sum += t;
            // return sum;
            // return accumulate(time_to_reach.begin(), time_to_reach.end());
            return max_time_to_reach_all;
        }

    }
};
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