<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>

struct DST {
    struct Node {
        uint head;
        uint rank;
    };
    Node* nodes;
    DST(uint n) {
        nodes = new Node[n];
        for (uint x = 0; x < n; x++) {
            nodes[x] = {x, 0};
        }
    }
    uint head(uint x) {
        uint head = x;
        for (;nodes[head].head != head; head = nodes[head].head);
        // for (;x != head;) {
        //     uint t = x;
        //     x = nodes[x].head;
        //     nodes[t].head = head;
        // }
        return head;
    }
    void merge(uint x, uint y) {
        x = head(x);
        y = head(y);
        if (nodes[x].rank < nodes[y].rank) {
            nodes[x].head = y;
        }
        else {
            nodes[y].head = x;
            if (nodes[x].rank == nodes[y].rank) {
                nodes[x].rank++;
            }
        }
    }
    bool same(uint x, uint y) {
        return head(x) == head(y);
    }
};

class Solution {
public:
    vector<bool> distanceLimitedPathsExist(int n, vector<vector<int>>& edgeList, vector<vector<int>>& queries2) {
        uint m = edgeList.size();
        vector<tuple<uint, uint, uint>> edges(m);
        for (uint x = 0; x < m; x++) {
            edges[x] = {edgeList[x][2], edgeList[x][0], edgeList[x][1]};
        }
        sort(edges.begin(), edges.end());
        DST dst(n);
        uint edgei = 0;
        uint q = queries2.size();
        vector<tuple<uint, uint, uint, uint>> queries(q);
        for (uint x = 0; x < q; x++) {
            queries[x] = {queries2[x][2],  queries2[x][0], queries2[x][1], x};
        }
        sort(queries.begin(), queries.end());
        vector<bool> ans(q);
        for (auto query : queries) {
            uint limit = get<0>(query);
            for (;edgei < m && get<0>(edges[edgei]) < limit; edgei++) {
                dst.merge(get<1>(edges[edgei]), get<2>(edges[edgei]));
            }
            ans[get<3>(query)] = dst.same(get<1>(query), get<2>(query));
        }
        return ans;
    }
};
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