zoukankan      html  css  js  c++  java
  • Geeks Union-Find Algorithm Union By Rank and Path Compression 图环算法

    相同是查找一个图是否有环的算法,可是这个算法非常牛逼,构造树的时候能够达到O(lgn)时间效率。n代表顶点数

    原因是依据须要缩减了树的高度,也叫压缩路径(Path compression),名字非常高深,只是事实上不难理解,简单来说就是每次查找一个节点的时候,都把这一路径中的全部节点都赋予根节点作为路径。


    原文没指出的地方:

    也由于须要压缩,所以初始化的时候注意,不能如前面简单有用Union Find的算法那样初始化全部顶点的父母节点为-1,应该是初始化全部节点的父母节点为本身(自己生殖自己?),然后就方便递归的时候一律能够返回这个跟节点了。 

    当然事实上初始化为-1也是能够的,只是须要额外代码处理一下,也不难。


    最后能够參考原文:http://www.geeksforgeeks.org/union-find-algorithm-set-2-union-by-rank/


    #pragma once
    #include <iostream>
    
    class UnionFindUnionByRank
    {
    	struct Edge
    	{
    		int src, des;
    	};
    
    	struct Graph
    	{
    		int V, E;
    		Edge *edges;
    		Graph(int v, int e) : V(v), E(e)
    		{
    			edges = new Edge[e];
    		}
    		~Graph()
    		{
    			if (edges) delete [] edges;
    		}
    	};
    
    	struct subSet
    	{
    		int parent, rank;
    	};
    
    	int find(subSet *subs, int i)
    	{
    		//由于要压缩,所以不能使用-1作为根的标志了
    		if (subs[i].parent != i)
    		{
    			//Union by rank: attach smaller rank tree to high rank tree. It is so simple, but very hard to create it totally by ourself, so it's good to stand on the shoulder of the giant.
    			subs[i].parent = find(subs, subs[i].parent);
    		}
    		return subs[i].parent;//由于假设-1作为根标志,那么这里就要返回i,就达不到压缩的效果了,而是应该返回parent,一层一层递归回上一层。
    	}
    
    	void unionTwo(subSet *subs, int x, int y)
    	{
    		int xroot = find(subs, x);
    		int yroot = find(subs, y);
    
    		if (subs[xroot].rank < subs[yroot].rank)
    		{
    			subs[xroot].parent = yroot;
    		}
    		else if (subs[xroot].rank > subs[yroot].rank)
    		{
    			subs[yroot].parent = xroot;
    		}
    		else
    		{
    			//only need to increment its rank when ther are equal rank
    			subs[yroot].parent = xroot;
    			subs[xroot].rank++;
    		}
    	}
    
    	bool isCycle(Graph *gra)
    	{
    		subSet *subs = new subSet[gra->V];
    		for (int i = 0; i < gra->V; i++)
    		{
    			subs[i].parent = i;//parent不能初始化为-1
    			subs[i].rank = 0;
    		}
    
    		for (int e = 0; e < gra->E; e++)
    		{
    			int x = find(subs, gra->edges[e].src);
    			int y = find(subs, gra->edges[e].des);
    
    			if (x == y) return true;
    
    			unionTwo(subs, x, y);
    		}
    
    		return false;
    	}
    public:
    	UnionFindUnionByRank()
    	{
    		int V = 3, E = 3;
    		struct Graph* graph = new Graph(V, E);
    
    		// add edge 0-1
    		graph->edges[0].src = 0;
    		graph->edges[0].des = 1;
    
    		// add edge 1-2
    		graph->edges[1].src = 1;
    		graph->edges[1].des = 2;
    
    		// add edge 0-2
    		graph->edges[2].src = 0;
    		graph->edges[2].des = 2;
    
    		if (isCycle(graph))
    			printf( "Union By Rank found graph contains cycle 
    " );
    		else
    			printf( "Union By Rank found graph doesn't contain cycle 
    " );
    	}
    };



  • 相关阅读:
    服务器操作nginx相关操作命令
    git使用命令
    超出隐藏显示
    微信小程序清除默认样式
    程序员提升之排查bug的能力
    call和apply的基本用法与区别
    vuejs 插件开发并发布到npm--(3)vue组件开发并发布
    vuejs 插件开发并发布到npm--(2)js插件开发
    vuejs 插件开发并发布到npm--(1)为什么要进行插件开发管理
    双机热备份和负载均衡的区别
  • 原文地址:https://www.cnblogs.com/gcczhongduan/p/3812742.html
Copyright © 2011-2022 走看看