“This is the 28th day of my participation in the Gwen Challenge in November. Check out the details: The Last Gwen Challenge in 2021.”

describe

We have n cities labeled from 1 to n. Two different cities with labels x and y are directly connected by a bidirectional road if and only if x and y share a common divisor strictly greater than some threshold. More formally, cities with labels x and y have a road between them if there exists an integer z such that all of the following are true:

  • x % z == 0,
  • y % z == 0, and
  • z > threshold.

Given the two integers, n and threshold, and an array of queries, you must determine for each queries[i] = [ai, bi] if cities ai and bi are connected directly or indirectly. (i.e. there is some path between them).

Return an array answer, where answer.length == queries.length and answer[i] is true if for the ith query, there is a path between ai and bi, or answer[i] is false if there is no path.

Example 1:

Input: n = 6, threshold = 2, queries = [[1,4],[2,5],[3,6]] Output: [false,false,true] Explanation: The divisors for each number: 1: 1 2: 1, 2 3: 1, 3 4: 1, 2, 4 5: 1, 5 6: 1, 2, 3, 6 Using the underlined divisors above the threshold, only cities 3 and 6 share a common divisor, so they are the only ones directly connected. The result of each query: 1 is not connected to 4 [2,5] 2 is not connected to 5 [3,6] 3 is connected to 6 through path 3--6Copy the code

Example 2:

Input: n = 6, threshold = 0, queries = [[4,5],[3,4],[3,2],[2,6],[1,3]]
Output: [true,true,true,true,true]
Explanation: The divisors for each number are the same as the previous example. However, since the threshold is 0,
all divisors can be used. Since all numbers share 1 as a divisor, all cities are connected.
Copy the code

Example 3:

Input: n = 5, threshold = 1, the queries = [[4, 5], [4, 5], [3, 2], [2, 3], [3, 4]] Output: [false,false,false,false,false] Explanation: Only cities 2 and 4 share a common divisor 2 which is strictly greater than the threshold 1, so they are the only ones directly connected. Please notice that there can be multiple queries for the same pair of nodes [x, y], and that the query [x, y] is equivalent to the query [y, x].Copy the code

Note:

  • 2 <= n <= 10^4
  • 0 <= threshold <= n
  • 1 <= queries.length <= 10^5
  • queries[i].length == 2
  • 1 <= ai, bi <= cities
  • ai! = bi

parsing

Given n cities, marked from 1 to n. If and only if x and Y share a common divisor strictly greater than a certain threshold, two different cities labeled X and Y are directly connected by a two-way road. If there is an integer z such that all of the following conditions are true, then there is a road between cities labeled X and Y:

  • x % z == 0
  • y % z == 0
  • z > threshold

Given two integers n and threshold and an array queries, for each queries[I] = [ai, bi], confirm whether city AI and BI are connected, whether directly or indirectly. Length == queries.length and answer[I] if there is a path between cities for the ith query, then answer[I] is true. If there is no path, then answer[I] is true. Then answer[I] is false.

In this case, the fast reading method is used to select two cities to merge. If the violent solution is used, the time complexity is too high to judge whether the two cities are connected. Another idea is to traverse the common divisor t greater than threshold, and then find multiples of T within the range less than or equal to N. As long as cities with multiples of T can be combined together, these cities can be given the smallest city ID as their ancestors. Finally, each pair of cities in queries is convenient. If their ancestors are the same, they are connected; otherwise, they are not connected.

answer

class Solution(object): def areConnected(self, n, threshold, queries): """ :type n: int :type threshold: int :type queries: List[List[int]] :rtype: List[bool] """ father = {i:i for i in range(1, n+1)} visited = {i:0 for i in range(1, n+1)} for t in range(threshold+1, n+1): if visited[t]:continue for x in range(t, n+1, t): visited[x] = 1 if self.getFather(x,father) ! = self.getFather(t,father): self.union(x, t,father) result = [] for query in queries: result.append(self.getFather(query[0],father)==self.getFather(query[1],father)) return result def union(self,a,b,father): a = father[a] b = father[b] if a<b: father[b] = a else: father[a] = b def getFather(self,x,father): if father[x] ! = x: father[x] = self.getFather(father[x],father) return father[x]Copy the code

The results

Given the linked links in the Python online submissions for Graph Connectivity With Threshold. Memory Usage: Given in Python online submissions for Graph Connectivity With Threshold.Copy the code

Original link: leetcode.com/problems/gr…

Your support is my biggest motivation