[LintCode] Find the Weak Connected Component in the Directed Graph
Find the number Weak Connected Component in the directed graph. Each node in the graph contains a label and a list of its neighbors. (a connected set of a directed graph is a subgraph in which any two vertices are connected by direct edge path.)
Given graph:
A----->B C
\ | |
\ | |
\ | |
\ v v
->D E <- F
Return {A,B,D}, {C,E,F}. Since there are two connected component which are{A,B,D} and {C,E,F}
Sort the element in the set in increasing order.
Solution:
并查集。遍历每一条变,更新每个节点所在集合。
/**
* Definition for Directed graph.
* struct DirectedGraphNode {
* int label;
* vector<DirectedGraphNode *> neighbors;
* DirectedGraphNode(int x) : label(x) {};
* };
*/
class Solution {
//use union-set to solve
private:
int find(unordered_map<int, int> &nodeMap, int label){
if(nodeMap.find(label) == nodeMap.end()){
nodeMap[label] = label;
return label;
//if this node doesn't belong to any union-set, create a new set
}else{
//this node belongs to some set, find the root of the set
int res = nodeMap[label];
while(nodeMap[res] != res)
res = nodeMap[res];
return res;
}
}
public:
/**
* @param nodes a array of directed graph node
* @return a connected set of a directed graph
*/
vector<vector<int>> connectedSet2(vector<DirectedGraphNode*>& nodes) {
unordered_map<int, int> nodeMap;
vector<vector<int> > result;
for(int i = ;i < nodes.size();++i){
for(int j = ;j < (nodes[i]->neighbors).size();++j){
int s1 = find(nodeMap, nodes[i]->label);
int s2 = find(nodeMap, (nodes[i]->neighbors)[j]->label);
if(s1 != s2){
//union two sets
if(s1 < s2) nodeMap[s2] = s1;
else nodeMap[s1] = s2;
}else{
//do nothing
}
}
} unordered_map<int, int> setId2VecId;
for(int i = ;i < nodes.size();++i){
int label = nodes[i]->label;
int setId = find(nodeMap, label);
if(setId2VecId.find(setId) == setId2VecId.end()){
vector<int> vec;
setId2VecId[setId] = result.size();
result.push_back(vec);
}
int idx = setId2VecId[setId];
result[idx].push_back(label);
}
for(int i = ;i < result.size();++i)
sort(result[i].begin(), result[i].end()); return result;
}
};
Inspired by: http://www.cnblogs.com/easonliu/p/4607300.html
[LintCode] Find the Weak Connected Component in the Directed Graph的更多相关文章
- Find the Weak Connected Component in the Directed Graph
Description Find the number Weak Connected Component in the directed graph. Each node in the graph c ...
- lintcode:Find the Connected Component in the Undirected Graph 找出无向图汇总的相连要素
题目: 找出无向图汇总的相连要素 请找出无向图中相连要素的个数. 图中的每个节点包含其邻居的 1 个标签和 1 个列表.(一个无向图的相连节点(或节点)是一个子图,其中任意两个顶点通过路径相连,且不与 ...
- [LintCode] Find the Connected Component in the Undirected Graph
Find the Connected Component in the Undirected Graph Find the number connected component in the undi ...
- LeetCode Number of Connected Components in an Undirected Graph
原题链接在这里:https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/ 题目: Giv ...
- LeetCode 323. Number of Connected Components in an Undirected Graph
原题链接在这里:https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/ 题目: Giv ...
- Connected Component in Undirected Graph
Description Find connected component in undirected graph. Each node in the graph contains a label an ...
- algorithm@ Strongly Connected Component
Strongly Connected Components A directed graph is strongly connected if there is a path between all ...
- [HDU6271]Master of Connected Component
[HDU6271]Master of Connected Component 题目大意: 给出两棵\(n(n\le10000)\)个结点的以\(1\)为根的树\(T_a,T_b\),和一个拥有\(m( ...
- Codeforces Round #575 (Div. 3) E. Connected Component on a Chessboard(思维,构造)
E. Connected Component on a Chessboard time limit per test2 seconds memory limit per test256 megabyt ...
随机推荐
- 繁华模拟赛 Vincent的城堡
#include<iostream> #include<cstdio> #include<string> #include<cstring> #incl ...
- 新浪微博的XSS漏洞攻击过程详解
今天晚上(2011年6月28日),新浪微博出现了一次比较大的XSS攻击事件.大量用户自动发送诸如:“郭美美事件的一些未注意到的细节”,“建 党大业中穿帮的地方”,“让女人心动的100句诗歌”,“3D肉 ...
- 第13章 使用Bind提供域名解析服务
章节简述: 本章节将让您理解DNS服务程序的原理,学习正向解析与反向解析实验,掌握DNS主服务器.从服务器.缓存服务器的部署方法. 够熟练配置区域信息文件与区域数据文件,以及通过使用分离解析技术让不同 ...
- [LA4108]SKYLINE
[LA4108]SKYLINE 试题描述 The skyline of Singapore as viewed from the Marina Promenade (shown on the left ...
- Python机器学习库scikit-learn实践
原文:http://blog.csdn.net/zouxy09/article/details/48903179 一.概述 机器学习算法在近几年大数据点燃的热火熏陶下已经变得被人所“熟知”,就算不懂得 ...
- Eclipse 项目红色叹号:Build Path Problem
Description Resource Path Location TypeA cycle was detected in the build path of project 'shgl-categ ...
- Java集合框架中List接口的简单使用
Java集合框架可以简单的理解为一种放置对象的容器,和数学中的集合概念类似,Java中的集合可以存放一系列对象的引用,也可以看做是数组的提升,Java集合类是一种工具类,只有相同类型的对象引用才可以放 ...
- HDU4870 Rating(概率)
第一场多校,感觉自己都跳去看坑自己的题目里去了,很多自己可能会比较擅长一点的题目没看,然后写一下其中一道概率题的题解吧,感觉和自己前几天做的概率dp的思路是一样的.下面先来看题意:一个人有两个TC的账 ...
- hdu 3032 Nim or not Nim? (SG函数博弈+打表找规律)
Nim or not Nim? Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Sub ...
- angularjs 指令(directive)详解(2)
原文地址 上一篇我们说到了transclude,那么,我们现在继续讲解之后的内容. 9.scope 可选参数,默认值为false.取值: false - 在这个directive里不会创建新的scop ...