<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>

/*You are given a network of n nodes, labeled from 1 to n. You are also given times, 
a list of travel times as directed edges times[i] = (ui, vi, wi), where ui is the 
source node, vi is the target node, and wi is the time it takes for a signal to 
travel from source to target.

We will send a signal from a given node k. Return the minimum time it takes for 
all the n nodes to receive the signal. If it is impossible for all the n nodes 
to receive the signal, return -1.
1 <= k <= n <= 100
1 <= times.length <= 6000
times[i].length == 3
1 <= ui, vi <= n
ui != vi
0 <= wi <= 100
All the pairs (ui, vi) are unique. (i.e., no multiple edges.)
*/
class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        //times[i] = (ui, vi, wi)
        vector<vector<int>> adj[n+1];
        for(auto& time: times) {
            int source = time[0];
            int dest = time[1];
            int travelTime = time[2];
            adj[ source ].push_back( {travelTime, dest}); //travletime, dest.
        };
        
        vector<int> signalReceivedAt(n+1, INT_MAX);
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq; //minimum
        pq.push({0, k}); //timeRecived, source.
        signalReceivedAt[k] = 0;
        while(!pq.empty()) {
            vector<int> curNode = pq.top();
            int curNodeTime = curNode[0];
            int curNodeNum = curNode[1];
            pq.pop();

            if(curNodeTime > signalReceivedAt[curNodeNum] ) //This node will not help.
                continue;
            for(vector<int> edge: adj[curNodeNum]) {
                int time = edge[0];
                int neighborNode = edge[1];
                if(signalReceivedAt[neighborNode] > curNodeTime + time ) {
                    signalReceivedAt[neighborNode] = curNodeTime + time;
                    pq.push({curNodeTime + time, neighborNode});
                }
            }
        }

        int ret = INT_MIN;
        for(int i=1; i<=n; i++) 
            ret = max(ret, signalReceivedAt[i]);
        return ret == INT_MAX? -1: ret;
    }
};
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