<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) {
        // create adj list
        vector<vector<vector<int>>> lists( n + 1 );
        for( const auto &t : times ) {
            // push target and time
            lists[ t[ 0 ] ].push_back( { t[ 1 ], t[ 2 ] } );
        }

        queue<vector<int>> q;
        q.push( { k, 0 } );
        unordered_map<int, int> vis;    // key: node, value: min time
        vis[k] = 0;

        while( !q.empty() ) {
            int q_num = q.size();

            for( int i = 0; i < q_num; i++ ) {
                // current node
                auto node = q.front();
                q.pop();
                int node_num = node[0];
                int cur_time = node[1];

                // travel to other nodes from node_num
                for( const auto &path : lists[node_num] ) {
                    int tar = path[0];
                    int time = path[1];

                    // push the visited node to queue with added time
                    // printf( "src: %d, tar: %d, curTime: %d, appTime: %d\n", node_num, tar, cur_time, time );
                    if( vis.count( tar ) == 0 || cur_time + time < vis[ tar ] ) {
                        // printf( "push: %d, %d\n", tar, cur_time + time );
                        vis[ tar ] = cur_time + time;
                        q.push( {tar, cur_time + time} );
                    }
                }
            }
        }

        if( vis.size() < n ) return -1;

        int nMaxTime = 0;
        for( auto it : vis ) {
            nMaxTime = max( nMaxTime, it.second );
        }

        return nMaxTime;
        // use bfs to traverse all nodes
        // when every node is reached,
        // 1. add to set
        // 2. calc the max of total sum of time, which is the answer

        // when the queue is empty
        // check if set size equals to n, return max of total sum of time if it is, or return -1
    }
};
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