dataStructure@ Check whether a given graph is Bipartite or not
Check whether a given graph is Bipartite or not
A Bipartite Graph is a graph whose vertices can be divided into two independent sets, U and V such that every edge (u, v) either connects a vertex from U to V or a vertex from V to U. In other words, for every edge (u, v), either u belongs to U and v to V, or u belongs to V and v to U. We can also say that there is no edge that connects vertices of same set.
A bipartite graph is possible if the graph coloring is possible using two colors such that vertices in a set are colored with the same color. Note that it is possible to color a cycle graph with even cycle using two colors. For example, see the following graph.
It is not possible to color a cycle graph with odd cycle using two colors.
Algorithm to check if a graph is Bipartite:
One approach is to check whether the graph is 2-colorable or not using backtracking algorithm m coloring problem.
Following is a simple algorithm to find out whether a given graph is Birpartite or not using Breadth First Search (BFS).
1. Assign RED color to the source vertex (putting into set U).
2. Color all the neighbors with BLUE color (putting into set V).
3. Color all neighbor’s neighbor with RED color (putting into set U).
4. This way, assign color to all vertices such that it satisfies all the constraints of m way coloring problem where m = 2.
5. While assigning colors, if we find a neighbor which is colored with same color as current vertex, then the graph cannot be colored with 2 vertices (or graph is not Bipartite)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<limits>
#include<vector>
#include<stack>
using namespace std;
struct edge{
int to, cost;
edge(int t){
this->to = t; this->cost = ;
}
};
void addEdge(vector<edge> &, vector<vector<int> > &, int, int);//add directed edge.
void buildMap(vector<edge> &edgelist, vector<vector<int> > &G){
addEdge(edgelist,G,,);
addEdge(edgelist,G,,);
addEdge(edgelist,G,,);
addEdge(edgelist,G,,);
addEdge(edgelist,G,,);
//addEdge(edgelist,G,5,0);
}
void addDoubleEdge(vector<edge> &, vector<vector<int> > &, int, int);// add undirected edge.
bool isCyclic(vector<edge>, vector<vector<int> >,vector<bool>, vector<bool>, int);// find cycles starting from v.
void isCyclicUtil(vector<edge>, vector<vector<int> >);// find all cycles.
bool dfs(vector<edge>, vector<vector<int> >, vector<bool>, int, int);//check if ''to'' is reachable from ''from''.
void isReachable(vector<edge>, vector<vector<int> >, int, int);
bool isBipartitie(vector<edge> , vector<vector<int> >,int v);//check if a graph is a bipartite graph.
int main(){
int maxn = ;
vector<edge> edgelist;
vector<vector<int> > G(maxn); buildMap(edgelist,G); //isCyclicUtil(edgelist, G); //isReachable(edgelist, G, 1, 1); if(isBipartitie(edgelist, G, )) cout<<"YES"<<endl;
else cout<<"NO"<<endl; return ;
}
bool isCyclic(vector<edge> edgelist, vector<vector<int> > G,vector<bool> vis, vector<bool> RecStack, int v){
for(int i=;i<G[v].size();++i){
edge e = edgelist[G[v][i]];
if(RecStack[e.to]) return true;
if(!vis[e.to]){
vis[e.to] = true; RecStack[e.to] = true;
if(isCyclic(edgelist,G,vis,RecStack,e.to)) return true;
RecStack[e.to] = false;
}
}
return false;
}
void isCyclicUtil(vector<edge> edgelist, vector<vector<int> > G){// find all cycles.
vector<bool> vis(G.size());
vector<bool> RecStack(G.size());
for(int i=;i<vis.size();++i) vis[i]=false;
for(int i=;i<RecStack.size();++i) RecStack[i]=false; for(int i=;i<G.size();++i){
if(!vis[i]){
vis[i] = true; RecStack[i] = true;
if(isCyclic(edgelist,G,vis,RecStack,i)){
cout<<i<<" starts a cycle"<<endl;
}
RecStack[i] = false;
}
}
}
void addEdge(vector<edge> &edgelist, vector<vector<int> > &G, int from, int to){
edgelist.push_back(edge(to));
G[from].push_back(edgelist.size()-);
}
void addDoubleEdge(vector<edge> &edgelist, vector<vector<int> > &G, int from, int to){
addEdge(edgelist,G,from,to);
addEdge(edgelist,G,to,from);
}
bool dfs(vector<edge> edgelist, vector<vector<int> > G, vector<bool> vis, int from, int to){
if(from == to) return true;
for(int i=;i<G[from].size();++i){
edge e = edgelist[G[from][i]];
if(e.to == to) return true;
if(!vis[e.to]){
vis[e.to] = true;
if(dfs(edgelist, G, vis, e.to, to)) return true;
}
}
return false;
}
void isReachable(vector<edge> edgelist, vector<vector<int> > G, int from, int to){
vector<bool> vis(G.size());
for(int i=;i<vis.size();++i) vis[i] = false;
vis[from] = true;
if(dfs(edgelist, G, vis, from, to)) cout<<from<<" and "<<to<<" are reachable to each other"<<endl;
else cout<<from<<" and "<<to<<" are not reachable to each other"<<endl;
}
bool isBipartitie(vector<edge> edgelist, vector<vector<int> > G,int v){
vector<int> color(G.size());
for(int i=;i<color.size();++i) color[i] = -;
stack<int> st;
while(!st.empty()) st.pop(); st.push(v); color[v]=;// 1 stands for RED, and 0 stands for BLUE, -1 stands for non-colored. while(!st.empty()){
int k = st.top(); st.pop(); for(int i=;i<G[k].size();++i){
edge e = edgelist[G[k][i]];
if(color[e.to] == -){
color[e.to] = - color[k];
st.push(e.to);
}
else if(color[e.to] == color[k]) return false;
}
}
return true;
}
dataStructure@ Check whether a given graph is Bipartite or not的更多相关文章
- dataStructure@ Check if a directed graph has cycles
#include<iostream> #include<cstdio> #include<cstring> #include<limits> #incl ...
- Geeks - Check whether a given graph is Bipartite or not 二分图检查
检查一个图是否是二分图的算法 使用的是宽度搜索: 1 初始化一个颜色记录数组 2 利用queue宽度遍历图 3 从随意源点出发.染色0. 或1 4 遍历这点的邻接点.假设没有染色就染色与这个源点相反的 ...
- LeetCode 785. Is Graph Bipartite?
原题链接在这里:https://leetcode.com/problems/is-graph-bipartite/ 题目: Given an undirected graph, return true ...
- Is Graph Bipartite?
Given an undirected graph, return true if and only if it is bipartite. Recall that a graph is bipart ...
- [LeetCode] Is Graph Bipartite? 是二分图么?
Given an undirected graph, return true if and only if it is bipartite. Recall that a graph is bipart ...
- [Swift]LeetCode785. 判断二分图 | Is Graph Bipartite?
Given an undirected graph, return true if and only if it is bipartite. Recall that a graph is bipart ...
- LeetCode - Is Graph Bipartite?
Given an undirected graph, return true if and only if it is bipartite. Recall that a graph is bipart ...
- [LeetCode] 785. Is Graph Bipartite?_Medium tag: DFS, BFS
Given an undirected graph, return true if and only if it is bipartite. Recall that a graph is bipart ...
- 785. Is Graph Bipartite?
Given an undirected graph, return true if and only if it is bipartite. Recall that a graph is bipart ...
随机推荐
- Samza在YARN上的启动过程 =》 之二 submitApplication
首先,来看怎么构造一个org.apache.hadoop.yarn.client.api.YarnClient class ClientHelper(conf: Configuration) exte ...
- [笨木头FireFly01]入门篇1·最简单的服务端和客户端连接
原地址:http://www.9miao.com/question-15-53938.html 最近一直在写游戏,几乎没有来写教程了,打算放慢一下脚步,学学新东西.那为嘛我要学FireFly呢? 之前 ...
- [LeetCode]Divide Two Integer
Divide two integers without using multiplication, division and mod operator. 思考:位运算.AC的时候真的想说一句“尼玛.. ...
- linux命令补全 忘记命令只记得开头
linux的shell不仅提供上下箭头来翻阅历史使用过的命令,还提供命令补全功能. 例如,你想创建一个文件夹,只记得是m开头的命令,此时可以: ①输入m ②按键盘上的Tab键两次 (有可能还出现这句话 ...
- Servlet课程0426(十)Servlet如何删除cookie
//如何删除Cookie案例 package com.tsinghua; import javax.servlet.http.*; import java.io.*; public class Coo ...
- ArcGIS学习记录—union、merge及append的区别
原文地址: ArcGIS问题:union.merge及append的主要区别[转] - Silent Dawn的日志 - 网易博客 http://gisman.blog.163.com/blog/st ...
- ZOJ1586——QS Network(最小生成树)
QS Network DescriptionIn the planet w-503 of galaxy cgb, there is a kind of intelligent creature nam ...
- linux 上下文切换带来的影响
1.保存CPU寄存器中的内容 2.CPU高速缓存中的内容失效 3.重新装载页表,用于给线程程安装一个新的虚拟地址空间,页表缓存失效
- BZOJ2870: 最长道路tree
题解: 子树分治的做法可以戳这里:http://blog.csdn.net/iamzky/article/details/41120733 可是码量... 这里介绍另一种好写又快的方法. 我们还是一颗 ...
- 判断String为空
1. et_string.equals("")|| et_string==null 2. ""用equal.null用== TextUtils.isEmpt ...