拓扑排序(Topological Sorting) 一.拓扑排序 含义 构造AOV网络全部顶点的拓扑有序序列的运算称为拓扑排序(Topological Sorting). 在图论中,拓扑排序(Topological Sorting)是一个有向无环图(DAG, Directed Acyclic Graph)的所有顶点的线性序列. 且该序列必须满足下面两个条件: ①每个顶点出现且只出现一次. ②若存在一条从顶点 A 到顶点 B 的路径,那么在序列中顶点 A 出现在顶点 B 的前面. 注:有向无环图(

Given an directed graph, a topological order of the graph nodes is defined as follow: For each directed edge A-->B in graph, A must before B in the order list. The first node in the order can be any node in the graph with no nodes direct to it. Find

Topological sorting/ ordering is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering. We can conduct topological sorting only on DAG (directed acyclic graph). Therefore, t

一.什么是拓扑排序 在图论中,拓扑排序(Topological Sorting)是一个有向无环图(DAG, Directed Acyclic Graph)的所有顶点的线性序列.且该序列必须满足下面两个条件: 每个顶点出现且只出现一次. 若存在一条从顶点 A 到顶点 B 的路径,那么在序列中顶点 A 出现在顶点 B 的前面. 有向无环图(DAG)才有拓扑排序,非DAG图没有拓扑排序一说. 例如,下面这个图: 它是一个 DAG 图,那么如何写出它的拓扑排序呢?这里说一种比较常用的方法: 从 DAG

定义: Topological Sorting is a method of arranging the vertices in a directed acyclic graph (DAG有向无环图) as a sequence, such that for every directed edge(u,v), 即u-> v, vertex u comes before v in the ordering 注意: A topological ordering is possible if and

DZY Loves Topological Sorting Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 866    Accepted Submission(s): 250 Problem Description A topological sort or topological ordering of a directed g

Topological Sorting Time limit: 1.0 secondMemory limit: 64 MB Michael wants to win the world championship in programming and decided to study N subjects (for convenience we will number these subjects from 1 to N). Michael has worked out a study plan

Find Function Optimization: After Path compression: int find(int x){ return root[x] == x ? x : (root[x] = find(root[x])); } Avoid Stack overflow: int find(int a){ while(root[a]!=a){ a=root[a]; } return a; }  Combined with rank : (Combined with the he

［抄题］: 现在你总共有 n 门课需要选,记为 0 到 n - 1.一些课程在修之前需要先修另外的一些课程,比如要学习课程 0 你需要先学习课程 1 ,表示为[0,1]给定n门课以及他们的先决条件,判断是否可能完成所有课程? 给定 n = 2,先决条件为 [[1,0]] 返回 true给定 n = 2,先决条件为 [[1,0],[0,1]] 返回 false ［思维问题］: 不知道为啥是图:抽象图,很多条边构造出来的图 ［一句话思路］: 拓扑排序:先统计度.边数,再把度=0的点放进queue,进

DZY Loves Topological Sorting Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1250    Accepted Submission(s): 403 Problem Description A topological sort or topological ordering of a directed g

DZY Loves Topological Sorting Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5195 Description A topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for ev

传送门 DZY Loves Topological Sorting Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 221    Accepted Submission(s): 52 Problem Description A topological sort or topological ordering of a directed

题意: 删去K条边,使拓扑排序后序列字典序最大 分析: 因为我们要求最后的拓扑序列字典序最大,所以一定要贪心地将标号越大的点越早入队.我们定义点i的入度为di. 假设当前还能删去k条边,那么我们一定会把当前还没入队的di≤k的最大的i找出来,把它的di条入边都删掉,然后加入拓扑序列. 删除的一定是小连大的边,因为大连小的边在拓扑序列生成的时候就去掉了 #include <iostream> #include <cstdio> #include <queue> #incl

传送门 题意简述:给出一张DAGDAGDAG,要求删去不超过kkk条边问最后拓扑序的最大字典序是多少. 思路:贪心帮当前不超过删边上限且权值最大的点删边,用线段树维护一下每个点的入度来支持查询即可. 注意要在选点的时候更新后继的入度 代码: #include<bits/stdc++.h> #define ri register int using namespace std; inline int read(){ int ans=0; char ch=getchar(); while(!isd

题目链接: hdu:http://acm.hdu.edu.cn/showproblem.php?pid=5195 bc(中文):http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=573&pid=1002 题解: 1.拓扑排序+贪心 #include<algorithm> #include<iostream> #include<cstring> #include<cstdi

graphdfsbfs 1.clone graph2.copy list with random pointer3.topological sorting4.permutations5.subsets6.n queens7.subsetsII 8.palindrome partitioning9.combination sum10.combination sumII11.word ladder 12.word ladderII 克隆图:先克隆点,再克隆边. 宽度优先搜索有模板,以后告诉了图中的一

1 Clone Graph   1  copy ervery nodes by bfs  2  add neighbors public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) { if (node == null) { return node; } List<UndirectedGraphNode> nodes = new ArrayList<>(); Map<UndirectedGraphNode,

1图论: 1.1  133. Clone Graph https://leetcode.com/problems/clone-graph/#/description 思路:这题可以对照拷贝随机链表那道题,首先拷贝图中的节点,然后拷贝每个结点的neighbors数组. 这题使用BFS,所有的BFS都可以看成是queue和unordered_map的组合实现的,该题模拟实现了一个queue,主要是因为后面要判断某个元素是否已经访问,所以使用一个vector来模拟操作,传统的queue会弹出元素,还需

There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of words from the dictionary, where words are sorted lexicographically by the rules of this new language. Derive the

Yet Another Source Code for LintCode Current Status : 232AC / 289ALL in Language C++, Up to date (2016-02-10) For more problems and solutions, you can see my LintCode repository. I'll keep updating for full summary and better solutions. See cnblogs t