- Difficulty: Hard
- Tags: LeetCode, Hard, Union Find, Graph, Minimum Spanning Tree, leetcode-1724, O(nlogn + m), 🔒, Versioned Union Find, Binary Lifting
Problem
An undirected graph of n nodes is defined by edgeList, where edgeList[i] = [ui, vi, disi] denotes an edge between nodes ui and vi with distance disi. Note that there may be multiple edges between two nodes, and the graph may not be connected.
Implement the DistanceLimitedPathsExist class:
DistanceLimitedPathsExist(int n, int[][] edgeList)Initializes the class with an undirected graph.boolean query(int p, int q, int limit)Returnstrueif there exists a path fromptoqsuch that each edge on the path has a distance strictly less thanlimit, and otherwisefalse.
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Example 1:
Input ["DistanceLimitedPathsExist", "query", "query", "query", "query"] [[6, [[0, 2, 4], [0, 3, 2], [1, 2, 3], [2, 3, 1], [4, 5, 5]]], [2, 3, 2], [1, 3, 3], [2, 0, 3], [0, 5, 6]] Output [null, true, false, true, false] Explanation DistanceLimitedPathsExist distanceLimitedPathsExist = new DistanceLimitedPathsExist(6, [[0, 2, 4], [0, 3, 2], [1, 2, 3], [2, 3, 1], [4, 5, 5]]); distanceLimitedPathsExist.query(2, 3, 2); // return true. There is an edge from 2 to 3 of distance 1, which is less than 2. distanceLimitedPathsExist.query(1, 3, 3); // return false. There is no way to go from 1 to 3 with distances strictly less than 3. distanceLimitedPathsExist.query(2, 0, 3); // return true. There is a way to go from 2 to 0 with distance < 3: travel from 2 to 3 to 0. distanceLimitedPathsExist.query(0, 5, 6); // return false. There are no paths from 0 to 5.
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Constraints:
2 <= n <= 1040 <= edgeList.length <= 104edgeList[i].length == 30 <= ui, vi, p, q <= n-1ui != vip != q1 <= disi, limit <= 109- At mostÂ
104calls will be made toquery.