<p>An undirected graph of <code>n</code> nodes is defined by <code>edgeList</code>, where <code>edgeList[i] = [u<sub>i</sub>, v<sub>i</sub>, dis<sub>i</sub>]</code> denotes an edge between nodes <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> with distance <code>dis<sub>i</sub></code>. Note that there may be <strong>multiple</strong> edges between two nodes.</p>

<p>Given an array <code>queries</code>, where <code>queries[j] = [p<sub>j</sub>, q<sub>j</sub>, limit<sub>j</sub>]</code>, your task is to determine for each <code>queries[j]</code> whether there is a path between <code>p<sub>j</sub></code> and <code>q<sub>j</sub></code><sub> </sub>such that each edge on the path has a distance <strong>strictly less than</strong> <code>limit<sub>j</sub></code> .</p>

<p>Return <em>a <strong>boolean array</strong> </em><code>answer</code><em>, where </em><code>answer.length == queries.length</code> <em>and the </em><code>j<sup>th</sup></code> <em>value of </em><code>answer</code> <em>is </em><code>true</code><em> if there is a path for </em><code>queries[j]</code><em> is </em><code>true</code><em>, and </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/08/h.png" style="width: 267px; height: 262px;" />
<pre>
<strong>Input:</strong> n = 3, edgeList = [[0,1,2],[1,2,4],[2,0,8],[1,0,16]], queries = [[0,1,2],[0,2,5]]
<strong>Output:</strong> [false,true]
<strong>Explanation:</strong> The above figure shows the given graph. Note that there are two overlapping edges between 0 and 1 with distances 2 and 16.
For the first query, between 0 and 1 there is no path where each distance is less than 2, thus we return false for this query.
For the second query, there is a path (0 -&gt; 1 -&gt; 2) of two edges with distances less than 5, thus we return true for this query.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/08/q.png" style="width: 390px; height: 358px;" />
<pre>
<strong>Input:</strong> n = 5, edgeList = [[0,1,10],[1,2,5],[2,3,9],[3,4,13]], queries = [[0,4,14],[1,4,13]]
<strong>Output:</strong> [true,false]
<strong>Explanation:</strong> The above figure shows the given graph.
</pre>

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

<ul>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= edgeList.length, queries.length &lt;= 10<sup>5</sup></code></li>
	<li><code>edgeList[i].length == 3</code></li>
	<li><code>queries[j].length == 3</code></li>
	<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub>, p<sub>j</sub>, q<sub>j</sub> &lt;= n - 1</code></li>
	<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>
	<li><code>p<sub>j</sub> != q<sub>j</sub></code></li>
	<li><code>1 &lt;= dis<sub>i</sub>, limit<sub>j</sub> &lt;= 10<sup>9</sup></code></li>
	<li>There may be <strong>multiple</strong> edges between two nodes.</li>
</ul>

#define ll int
class Solution {
    ll findpar(ll node,vector<ll>& par){
        if(node==par[node]){
            return node;
        }
        return par[node]=findpar(par[node],par);
    }
    void unite(ll u,ll v,vector<ll>& par){
        u=findpar(u,par);
        v=findpar(v,par);
        if(u==v){
            return;
        }
        par[u]=v;
    }
public:
    vector<bool> distanceLimitedPathsExist(int n, vector<vector<int>>& edgeList, vector<vector<int>>& queries) {
        vector<ll> par(n,0);
        for(ll i=0;i<n;i++){
            par[i]=i;
        }
        priority_queue<pair<int,pair<int,int>>,vector<pair<int,pair<int,int>>>,greater<pair<int,pair<int,int>>>> pq;
        for(ll i=0;i<edgeList.size();i++){
            pq.push({edgeList[i][2],{edgeList[i][0],edgeList[i][1]}});
        }
        //priority_queue<pair<int,pair<int,int>>,vector<pair<int,pair<int,int>>>,greater<pair<int,pair<int,int>>>> pq;
        // sort(edgeList.begin(),edgeList.end());
        // for(int i=0;i<queries.size();i++){
        //     reverse(queries[i].begin(),queries[i].end());
        //     queries[i].push_back(i);
        // }
        priority_queue<pair<int,vector<int>>,vector<pair<int,vector<int>>>,greater<pair<int,vector<int>>>> q;
        for(int i=0;i<queries.size();i++){
            q.push({queries[i][2],{queries[i][0],queries[i][1],i}});
        }
        vector<bool> ans(queries.size());
        while(!q.empty()){
            auto it=q.top();
            q.pop();
            ll limit=it.first;
            ll x=it.second[0];
            ll y=it.second[1];
            ll ind=it.second[2];
            // ll z=edgeList[start][0];
            while(!pq.empty() && pq.top().first<limit){
                unite(pq.top().second.first,pq.top().second.second,par);
                pq.pop();
            }
            if(findpar(x,par)==findpar(y,par)){
                ans[ind]=true;
            }
            else{
                ans[ind]=false;
            }

        }
        return ans;


    }
};
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