<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)
    {
        k = k-1;
        const auto getFrom = [](const std::vector<int>& v) -> int { return v[0] -1; };
        const auto getTo = [](const std::vector<int>& v) -> int { return v[1] -1; };
        const auto getCost = [](const std::vector<int>& v) -> int { return v[2]; };
        
        using NeighborAndCost = std::pair<std::size_t, int>;
        std::vector<std::vector<NeighborAndCost>> fromTo(n);
        for(const std::vector<int>& edge : times)
            fromTo[getFrom(edge)].emplace_back(getTo(edge), getCost(edge));

        static const int UNREACHABLE = std::numeric_limits<int>::max();
        std::vector<int> reachCost(n, UNREACHABLE);
        reachCost[k] = 0;

        for(std::size_t i = 0; i < n; ++i)
        {
            for(std::size_t node = 0; node < n; ++node)
            {
                for(auto [neighbor, cost] : fromTo[node])
                {
                    if(reachCost[node] != UNREACHABLE)
                        reachCost[neighbor] = std::min(reachCost[neighbor], reachCost[node] + cost);
                }
            }
        }

        int delayTime = *std::max_element(std::begin(reachCost), std::end(reachCost));
        return delayTime == UNREACHABLE ? -1 : delayTime;
    }
};
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