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

        int minCities = edges.size();
        int minCity = -1;

        for (int i = 0; i < nodes.size(); i++) {
            int reachableCities = dijkstraSearch(nodes,i,distanceThreshold,minCities);
            // As we are incrementing i, if reachable == minCities we always replace minCity with i because i will always be bigger.
            if (reachableCities <= minCities) {
                minCities = reachableCities;
                minCity = i;
            }
        }

        return minCity;
    }
private:
#define USE_VECTOR_VISITED 1
    int dijkstraSearch(std::vector<std::vector<std::pair<int, int>>>& nodes, const int startCity, const int distanceThreshold, const int currentMinCities) {
        std::priority_queue<std::pair<int, std::pair<int,int>>, std::vector<std::pair<int, std::pair<int, int>>>, std::greater<std::pair<int, std::pair<int, int>>>> q;
        q.push(std::pair<int, std::pair<int,int>>(0, std::pair<int,int>(startCity,-1)));
        int reachable = -1;
#if USE_VECTOR_VISITED
        std::vector<bool> visited(nodes.size());
#else
        std::unordered_set<int> visitedSet;
#endif
        while (!q.empty()) {
            int thisNode = q.top().second.first;
#if USE_VECTOR_VISITED
            if (visited[thisNode]) {
#else
            if (visitedSet.contains(thisNode)) {
#endif
                q.pop();
                continue;
            }
            int accumulatedDistance = q.top().first;
            int parentNode = q.top().second.second;
#if USE_VECTOR_VISITED
            visited[thisNode] = true;
#else
            visitedSet.insert(thisNode);
#endif
            reachable++;
            if (reachable > currentMinCities) {
                // No point looking any further, even though the answer is wrong
                return reachable;
            }
            q.pop();
            for (int j = 0; j < nodes[thisNode].size(); j++) {
#if USE_VECTOR_VISITED
                if(!visited[nodes[thisNode][j].first]) {
#else
                if (!visitedSet.contains(nodes[thisNode][j].first)) {
#endif
                    int nextDistance = accumulatedDistance + nodes[thisNode][j].second;
                    if (nextDistance > distanceThreshold) {
                        continue;
                    }
                    q.push(std::pair<int, std::pair<int, int>>(nextDistance, std::pair<int, int>(nodes[thisNode][j].first, thisNode)));

                }
            }
        }
        return reachable;
    }
};
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