<p>There are <code>n</code> cities numbered from <code>0</code> to <code>n-1</code>. Given the array <code>edges</code> where <code>edges[i] = [from<sub>i</sub>, to<sub>i</sub>, weight<sub>i</sub>]</code> represents a bidirectional and weighted edge between cities <code>from<sub>i</sub></code> and <code>to<sub>i</sub></code>, and given the integer <code>distanceThreshold</code>.</p>

<p>Return the city with the smallest number of cities that are reachable through some path and whose distance is <strong>at most</strong> <code>distanceThreshold</code>, If there are multiple such cities, return the city with the greatest number.</p>

<p>Notice that the distance of a path connecting cities <em><strong>i</strong></em> and <em><strong>j</strong></em> is equal to the sum of the edges&#39; weights along that path.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/08/23/problem1334example1.png" style="width: 300px; height: 224px;" /></p>

<pre>
<strong>Input:</strong> n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4
<strong>Output:</strong> 3
<strong>Explanation: </strong>The figure above describes the graph.&nbsp;
The neighboring cities at a distanceThreshold = 4 for each city are:
City 0 -&gt; [City 1, City 2]&nbsp;
City 1 -&gt; [City 0, City 2, City 3]&nbsp;
City 2 -&gt; [City 0, City 1, City 3]&nbsp;
City 3 -&gt; [City 1, City 2]&nbsp;
Cities 0 and 3 have 2 neighboring cities at a distanceThreshold = 4, but we have to return city 3 since it has the greatest number.
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/08/23/problem1334example0.png" style="width: 300px; height: 224px;" /></p>

<pre>
<strong>Input:</strong> n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2
<strong>Output:</strong> 0
<strong>Explanation: </strong>The figure above describes the graph.&nbsp;
The neighboring cities at a distanceThreshold = 2 for each city are:
City 0 -&gt; [City 1]&nbsp;
City 1 -&gt; [City 0, City 4]&nbsp;
City 2 -&gt; [City 3, City 4]&nbsp;
City 3 -&gt; [City 2, City 4]
City 4 -&gt; [City 1, City 2, City 3]&nbsp;
The city 0 has 1 neighboring city at a distanceThreshold = 2.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 100</code></li>
	<li><code>1 &lt;= edges.length &lt;= n * (n - 1) / 2</code></li>
	<li><code>edges[i].length == 3</code></li>
	<li><code>0 &lt;= from<sub>i</sub> &lt; to<sub>i</sub> &lt; n</code></li>
	<li><code>1 &lt;= weight<sub>i</sub>,&nbsp;distanceThreshold &lt;= 10^4</code></li>
	<li>All pairs <code>(from<sub>i</sub>, to<sub>i</sub>)</code> are distinct.</li>
</ul>

class Solution {
public:
    int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {

        std::vector<std::vector<std::pair<int, int>>> nodes(n);
        for (auto& v : edges) {
            nodes[v[0]].push_back({v[1], v[2]});
            nodes[v[1]].push_back({v[0], v[2]});
        }

        int best_result = INT_MIN;
        int city_count = INT_MAX;

        auto cmp = [](std::pair<int, int>& left, std::pair<int, int>& right) { return left.second > right.second; };
         
        
        std::vector<bool> closed_set(100);

        for (int city = 0; city < n;city++) {
            std::priority_queue<std::pair<int, int>,std::vector<std::pair<int, int>>,decltype(cmp)> open_list;
            open_list.push({city, 0});

            
            //for(int i = 0; i < 100;i++) closed_set[i] = false;
            std::fill(closed_set.begin(),closed_set.end(),false);

            while (!open_list.empty()) {
                auto& node = open_list.top();
                int current_node = node.first;
                int current_dist = node.second;

                open_list.pop();
                closed_set[current_node] = true;

                for (auto& n : nodes[current_node]) {
                    if (current_dist + n.second <= distanceThreshold && !closed_set[n.first]) {
                        open_list.push({n.first, current_dist + n.second});
                    }
                }
            }

            int nodes_in_range = 0;
            for(int i = 0; i < 100;i++) if(closed_set[i]) nodes_in_range++;

            if (city_count > nodes_in_range) {
                best_result = city;
                city_count = nodes_in_range;
            } else if (city_count == nodes_in_range &&
                       best_result < city) {
                best_result = city;
            }
            
        }

        return best_result;
    }
};
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