<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 bfs(vector<vector<int>>& times, int n, int k) {
        // create adjacency list
        vector<vector<int>> adj[n+1];
        for (int i=0; i<times.size(); i++) {
            int u = times[i][0];    // from
            int v = times[i][1];    // to
            int w = times[i][2];    // weight

            adj[u].push_back({v,w});
        }

        int time = 0;

        vector<int> visited(n+1, 0);
        vector<int> distance(n+1, INT_MAX);

        priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> bfsq;
        bfsq.push({0,k});
        visited[k] = 1;
        distance[k] = 0;

        while (!bfsq.empty()) {
            // int max_time=0;
            int front_time = bfsq.top().first;
            int front_node = bfsq.top().second;

            bfsq.pop();

            for (int i=0; i<adj[front_node].size(); i++) {
                int ngbr = adj[front_node][i][0];
                int dist = adj[front_node][i][1];

                // if (!visited[ngbr]) {
                    if (front_time + dist < distance[ngbr]) {
                        distance[ngbr] = front_time + dist;
                        bfsq.push({distance[ngbr], ngbr});
                    }
                    // visited[ngbr] = 1;
                // }
            }
            // time += max_time;
        }
        int min_time = 0;
        for (int i=1; i<=n; i++) {
            if (distance[i]>min_time)   min_time = distance[i];
        }
        return min_time;
    }

    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        int time = bfs(times, n, k);
        if (time==INT_MAX)  return -1;
        return time;
    }
};
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