<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:
    vector<pair<int,int>>vec[102];
    int dist[102];


    int dijk(int src,int dt,int n)
    {
        dist[src]=0;
        pair<int,int> p;
        set<pair<int,int>>s;
        p.first=0;
        p.second=src;
        s.insert(p);  

        while(!s.empty())
        {
            p=*s.begin();
            int u=p.second;
            s.erase(p);
            for(int i=0;i<vec[u].size();i++)
            {
                p=vec[u][i];
                int v=p.first,d=p.second;
                //printf("%d %d %d\n",u,v,d);
                if(dist[v]==INT_MAX)
                {
                    dist[v]=dist[u]+d;
                    p.first=dist[v];
                    p.second=v;
                    s.insert(p);
                }
                else if(dist[u]+d<dist[v])
                {
                    p.first=dist[v];
                    p.second=v;
                    s.erase(p);
                    p.first=dist[u]+d;
                    s.insert(p);
                    dist[v]=dist[u]+d;
                }
            }
        }    
        int c=0;
        for(int i=0;i<n;i++)
        {
            printf("%d ",dist[i]);
            if(dist[i]<=dt)
                c++;
        }  
        printf("\n%d %d\n",src,c);
        return c;
    }

    int findTheCity(int n, vector<vector<int>>& arr, int dt) {
        for(int i=0;i<n;i++)
        {
            vec[i].clear();
        }
        pair<int,int> p;
        pair<int,pair<int,int>> pp;
        for(int i=0;i<arr.size();i++)
        {
            int u=arr[i][0],v=arr[i][1],d=arr[i][2];
            p.first=v;
            p.second=d;
            vec[u].push_back(p);
            p.first=u;
            vec[v].push_back(p);
            printf("%d %d %d\n",u,v,d);
        }
        int ans=INT_MAX,node=-1,x;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<=n;j++)
            {
                dist[j]=INT_MAX;
            }
             x=dijk(i,dt,n);
            if(x<=ans)
            {
                ans=x;
                node=i;
            }
        }
        return node;
    }
};
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