<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>

using Node = std::tuple<int, int, int, int>;

class Compare {
    public:
        bool operator()(const Node & L, const Node & R){
            //this is a min heap that prioritizes minimal target cost
            return !(std::get<3>(L) <= std::get<3>(R));
        }
};

class Solution {
public:

    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        //turn the times array into a map for easier lookup
        //node -> (target_node, cost)
        std::unordered_multimap<int, std::pair<int, int>> mmap;
        for(auto time : times) {
            mmap.emplace(time[0], std::pair(time[1], time[2]));
        }

        //djikstras
        //current node, target node, current cost, target cost
        std::priority_queue<Node, std::vector<Node>, Compare> pq;
        int number_of_visited_nodes = 0;
        std::vector<bool> visited(n, false);

        int max_cost = 0;

        pq.push({k, k, 0, 0});
        // visited[k] = true;
   
        while(!pq.empty()){
            
       
            
            
            auto entry = pq.top();
            std::cout << std::get<3>(entry) << std::endl;
            pq.pop();
            //check all the children of k
            int node = std::get<1>(entry);
            //check all the children
            if(visited[node]){
                continue;
            }
            
            number_of_visited_nodes++;
            visited[node] = true;
            //not visited
            max_cost = std::max(max_cost, std::get<3>(entry));
            //check all the children
            auto it_range = mmap.equal_range(node);
            for(auto it = it_range.first; it != it_range.second; it++){
                auto children = it -> second;
                pq.push({node, children.first, std::get<3>(entry), std::get<3>(entry) + children.second});
            }
        }
        std::cout << number_of_visited_nodes << std::endl;
        if(number_of_visited_nodes == n){
            return max_cost;
        } 

        
        return -1;
    }
};
56265