之前我已经写过一期的并查集的内容，这里主要是实现并查集的整体结构：

#pragma once
//并查集：class UnionFindSet//父母表示法;
{
public://构造函数：UnionFindSet(size_t size=0):_ufs(size,-1){}//合并元素;bool Union(int x, int y){//找到两组元素的rootint root1 = FindRoot(x);int root2 = FindRoot(y);//检查：//1.如果两组的root相同，返回false//2.如果两组的root不同，合并，返回trueif (root1 == root2){return false;}else{_ufs[root1] += _ufs[root2];_ufs[root2] = root1;//这里也可以通过压缩路径的方式，简化return true;}}bool UnionSize(int x, int y){int root1 = FindRoot(x);int root2 = FindRoot(y);if (x == y){return false;}else{//按照size合并，数量少的的合并数量多的int size1 = SizeRoot(root1);int size2 = SizeRoot(root2);if (size1 < size2){swap(root1, root2);}_ufs[root1] += _ufs[root2];_ufs[root2] = root1;		return true;}}//查找一个合并集合的root：//非递归写法：int FindRoot(int x){//查找root（<0)while (_ufs[x] > 0){x = _ufs[x];}return _ufs[x];}//递归写法：/*int FindRoot(int x){if (x == _ufs[x])return x;elsereturn FindRoot(_ufs[x]);}*///树的个数:size_t Count()const{size_t count = 0;for (auto e : _ufs){count += (e < 0 ? 1 : 0);}return count;//return x == fa[x] ? x : (fa[x] == findset2(fa[x]));//简写}//路径压缩：//递归版;int Path_compression(int x){if (x == _ufs[x])return x;_ufs[x] = Path_compression(_ufs[x]);return _ufs[x];}//非递归版;/*int Path_compression(int x){int root = FindRoot(root);while(_ufs[x] != root){int y = _ufs[x];_ufs[x] = root;x = y;}return root;}*///一个树的元素个数：int SizeRoot(int x){int root = FindRoot(x);return abs(_ufs[root]);}
private:vector<int> _ufs;
};

最后，祝福大家学习进步！！！