方式2,激进的压缩方式:find(x){while(x!=parent[x]) { parent[x]=find(parent[x]); x=parent[x]; }returnx; },此过程使得压缩后指定节点及其各非根父节点均直接成为根节点的子节点(同理想象下上述单链例子的压缩过程),在一些场景下该法很有用(见后文除法求值的实例),与上一种相比缺点是单节点压缩...
Weighted QuickUnionFind:Construct: O(n); Find: O(logn); Union:O(logn) Weighted QuickUnionFind with Path Compression: Construct: O(n); Find: amortizedO(1); Union:amortizedO(1) leetcode里使用UnionFind的题主要有:Number of Islands(lc200), LongestConsecutiveSequence(lc128), SurroundedRegion(lc...
Given an unsorted array of integers, find the length of the longest consecutive elements sequence. Your algorithm should run in O(n) complexity. Example: Input: [100, 4, 200, 1, 3, 2] Output: 4 Explanation: The longest consecutive elements sequence is[1, 2, 3, 4]. Therefore its leng...
而“我的 JavaScript 比你的 Rust 更快”的结论也是来自这个打赌。
建议和leetcode 685. Redundant Connection II 并查集Union Find 一起学习 代码如下: #include <iostream> #include <vector> #include <map> #include <set> #include <queue> #include <stack> #include <string> #include <climits> #include <algorithm> ...
现在我们的 Union-Find 算法主要需要实现这两个API: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 classUF{/* 将 p 和 q 连接 */publicvoidunion(int p,int q);/* 判断 p 和 q 是否连通 */publicbooleanconnected(int p,int q);/* 返回图中有多少个连通分量 */publicintcount();} ...
Union Find Princeton's lecture note on Union FindinAlgorithms and Data StructuresIt is a well organized note with clear illustration describing from the naive QuickFind to the one with Weighting and Path compression. With Weighting and Path compression, The algorithm runs in ...
Union-Find算法应用 上篇文章很多读者对于 Union-Find 算法的应用表示很感兴趣,这篇文章就拿几道 LeetCode 题目来讲讲这个算法的巧妙用法。 首先,复习一下,Union-Find 算法解决的是图的动态连通性问题,这个算法本身不难,能不能应用出来主要是看你抽象问题的能力,是否能够把原始问题抽象成一个有关图论的...
pythonmachine-learningocrsvmsupport-vector-machinefinal-year-projectsignature-verificationunion-findocr-recognitionconnected-componentsline-sweep-algorithmcapstone-projectsignature-detection UpdatedFeb 22, 2025 Python Set of Patterns to solve many algorithmic questions of similar type on LeetCode ...
#include <algorithm> using namespace std; const int maxx=10010; int pre[maxx]; int find(int x){ if(pre[x]!=x) pre[x]=find(pre[x]); return pre[x]; } int main (){ for(int i=0;i<maxx;i++) pre[i]=i; int n,m; cin>>n>>m; int com,x,y; while(m--){ cin>>com...