<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:
    typedef pair<int ,int> P;
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        priority_queue<P, vector<P>, greater<P> > pq;
        unordered_map<int, vector<vector<int> > > adj;
        for(auto &edge: times){
            int u = edge[0];
            int v = edge[1];
            int w = edge[2];
            adj[u].push_back({v, w});
        }

        vector<int> sDist(n+1);
        sDist[0] = -1;
        for(int i=1; i<n+1; i++){
            sDist[i] = INT_MAX;
        }
        sDist[k] = 0;
        pq.push({0, k});
        while(!pq.empty()){
            P front = pq.top();
            pq.pop();
            int dist = front.first;
            int node = front.second;
            for(auto &it: adj[node]){
                int dest = it[0];
                int totalDist = it[1] + dist;
                if(totalDist < sDist[dest]){
                    sDist[dest] = totalDist;
                    pq.push({totalDist, dest});
                }
            }
        }

        int maxi = INT_MIN;
        for(int i=1; i<n+1; i++){
            maxi = max(maxi, sDist[i]);
        }
        if(maxi == INT_MAX){
            return -1;
        }
        return maxi;
    }
};
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