zoukankan      html  css  js  c++  java
  • dataStructure@ Check whether a given graph is Bipartite or not

    Check whether a given graph is Bipartite or not

    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.

    Bipartite1

    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.

    Bipartite2

    It is not possible to color a cycle graph with odd cycle using two colors.
    Bipartite3

    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 = 0;
        }
    };
    void addEdge(vector<edge> &, vector<vector<int> > &, int, int);//add directed edge.
    void buildMap(vector<edge> &edgelist, vector<vector<int> > &G){
        addEdge(edgelist,G,0,1);
        addEdge(edgelist,G,1,2);
        addEdge(edgelist,G,2,3);
        addEdge(edgelist,G,3,4);
        addEdge(edgelist,G,4,0);
        //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 = 5;
        vector<edge> edgelist;
        vector<vector<int> > G(maxn);
        
        buildMap(edgelist,G);
        
        //isCyclicUtil(edgelist, G);
        
        //isReachable(edgelist, G, 1, 1);
        
        if(isBipartitie(edgelist, G, 0)) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
        
        return 0;
    }
    bool isCyclic(vector<edge> edgelist, vector<vector<int> > G,vector<bool> vis, vector<bool> RecStack, int v){
        for(int i=0;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=0;i<vis.size();++i) vis[i]=false;
        for(int i=0;i<RecStack.size();++i) RecStack[i]=false;
        
        for(int i=0;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()-1);
    }
    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=0;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=0;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=0;i<color.size();++i) color[i] = -1;
        stack<int> st; 
        while(!st.empty()) st.pop();
        
        st.push(v); color[v]=1;// 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=0;i<G[k].size();++i){
                edge e = edgelist[G[k][i]];
                if(color[e.to] == -1){
                    color[e.to] = 1 - color[k];
                    st.push(e.to);
                } 
                else if(color[e.to] == color[k]) return false;
            }
        }
        return true;
    }
    View Code
  • 相关阅读:
    再战设计模式(九)之组合模式
    再战设计模式(八)之桥接模式
    再战设计模式(七)之代理模式
    nyoj 题目2 括号配对问题
    剑指offer 面试题38
    杭电 1005
    九度oj 题目1552:座位问题
    九度oj 题目1482:玛雅人的密码 清华大学机试
    九度oj 题目1496:数列区间
    九度oj 题目1495:关键点
  • 原文地址:https://www.cnblogs.com/fu11211129/p/4906281.html
Copyright © 2011-2022 走看看