<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:
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        unordered_map<int, vector<pair<int, int>>> edges;
        for (const auto& time : times) {
            edges[time[0]].emplace_back(time[1], time[2]);
        }
        // create adjacent map 0 is pointer node 1 is cost and 2 is pointed node

        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<>>
            minHeap; // used to store weight and node
        minHeap.push({0, k}); // at cost 0 it takes to get to k node

        set<int> visited;
        int t = 0;
        // bfs search
        while (!minHeap.empty()) {
            auto curr = minHeap.top();
            minHeap.pop();
            int w1 = curr.first, n1 = curr.second;
            if (visited.count(n1)) { // if seen skip
                continue;
            }
            visited.insert(n1); 
            t = w1; // current weight

            if (edges.count(n1)) { // if has values we go through
                for (const auto& next : edges[n1]) {
                    int n2 = next.first, w2 = next.second;
                    if (!visited.count(n2)) {
                        minHeap.push({w1 + w2, n2});
                    }
                }
            }
        }

        if (visited.size() == n) {
            return t;
        } else {
            return -1;
        }
    }
};

45003