<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 {

using EDGE = vector<int>;

public:
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        vector<vector<EDGE>> srcEdgeArr;
        srcEdgeArr.resize(n + 1);
        for (auto time : times) {
            int src = time[0];
            int dst = time[1];
            int wgt = time[2];
            srcEdgeArr[src].push_back({wgt, src, dst});
        }
        priority_queue<EDGE, vector<EDGE>, greater<>> costPQ;
        unordered_set<int> receivedNodeSet;
        for (auto edge : srcEdgeArr[k]) {
            costPQ.push(edge);
        }
        receivedNodeSet.insert(k);
        int res = INT_MIN;
        while (!costPQ.empty()) {
            int cost = costPQ.top()[0];
            int src = costPQ.top()[1];
            int dst = costPQ.top()[2];
            costPQ.pop();

            if (receivedNodeSet.count(dst)) {
                continue;
            } else {
                receivedNodeSet.insert(dst);
            }

            res = max(res, cost);
            if (receivedNodeSet.size() == n) {
                return res;
            }

            for (auto edge : srcEdgeArr[dst]) {
                if (receivedNodeSet.count(edge[2])) {
                    continue;
                } else {
                    costPQ.push({cost + edge[0], edge[1], edge[2]});
                }
            }
        }
        return -1;
    }
};

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