zoukankan      html  css  js  c++  java
  • dataStructure@ Find if there is a path between two vertices in a directed graph

    Given a Directed Graph and two vertices in it, check whether there is a path from the first given vertex to second. For example, in the following graph, there is a path from vertex 1 to 3. As another example, there is no path from 3 to 0.

    We can either use Breadth First Search (BFS) or Depth First Search (DFS) to find path between two vertices. Take the first vertex as source in BFS (or DFS), follow the standard BFS (or DFS). If we see the second vertex in our traversal, then return true. Else return false.

    Following is C++ code that uses BFS for finding reachability of second vertex from first vertex.

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<limits>
     5 #include<vector>
     6 using namespace std;
     7 const int maxn = 4;
     8 struct edge{
     9     int to, cost;
    10     edge(int t){
    11         this->to = t; this->cost = 0;
    12     }
    13 };
    14 void addEdge(vector<edge> &edgelist, vector<vector<int> > &G, int from, int to){
    15     edgelist.push_back(edge(to));
    16     G[from].push_back(edgelist.size()-1);
    17 }
    18 void addDoubleEdge(vector<edge> &edgelist, vector<vector<int> > &G, int from, int to){
    19     addEdge(edgelist,G,from,to);
    20     addEdge(edgelist,G,to,from);
    21 }
    22 bool isCyclic(vector<edge> edgelist, vector<vector<int> > G,vector<bool> vis, vector<bool> RecStack, int v){
    23     for(int i=0;i<G[v].size();++i){
    24         edge e = edgelist[G[v][i]];
    25         if(RecStack[e.to]) return true;
    26         if(!vis[e.to]){
    27             vis[e.to] = true; RecStack[e.to] = true;
    28             if(isCyclic(edgelist,G,vis,RecStack,e.to)) return true;
    29             RecStack[e.to] = false;
    30         }
    31     }
    32     return false;
    33 }
    34 void isCyclicUtil(vector<edge> edgelist, vector<vector<int> > G){// find all cycles.
    35     vector<bool> vis(G.size());
    36     vector<bool> RecStack(G.size());
    37     for(int i=0;i<vis.size();++i) vis[i]=false;
    38     for(int i=0;i<RecStack.size();++i) RecStack[i]=false;
    39     
    40     for(int i=0;i<G.size();++i){
    41         if(!vis[i]){
    42             vis[i] = true; RecStack[i] = true;
    43             if(isCyclic(edgelist,G,vis,RecStack,i)){
    44                 cout<<i<<" starts a cycle"<<endl; 
    45             }
    46             RecStack[i] = false;
    47         }
    48     }
    49 }
    50 bool dfs(vector<edge> edgelist, vector<vector<int> > G, vector<bool> vis, int from, int to){
    51     if(from == to) return true;
    52     for(int i=0;i<G[from].size();++i){
    53         edge e = edgelist[G[from][i]];
    54         if(e.to == to) return true;
    55         if(!vis[e.to]){
    56             vis[e.to] = true;
    57             if(dfs(edgelist, G, vis, e.to, to)) return true;
    58         }
    59     }
    60     return false;
    61 }
    62 void isReachable(vector<edge> edgelist, vector<vector<int> > G, int from, int to){
    63     vector<bool> vis(G.size());
    64     for(int i=0;i<vis.size();++i) vis[i] = false;
    65     vis[from] = true;
    66     if(dfs(edgelist, G, vis, from, to)) cout<<from<<" and "<<to<<" are reachable to each other"<<endl;
    67     else cout<<from<<" and "<<to<<" are not reachable to each other"<<endl;
    68 }
    69 void buildMap(vector<edge> &edgelist, vector<vector<int> > &G){
    70     addEdge(edgelist,G,0,1);
    71     addEdge(edgelist,G,0,2);
    72     addEdge(edgelist,G,2,0);
    73     addEdge(edgelist,G,1,2);
    74     addEdge(edgelist,G,2,3);
    75     addEdge(edgelist,G,3,3);
    76 }
    77 int main(){
    78     vector<edge> edgelist;
    79     vector<vector<int> > G(maxn);
    80     
    81     buildMap(edgelist,G);
    82     
    83     isCyclicUtil(edgelist, G);
    84     
    85     isReachable(edgelist, G, 1, 1);
    86     
    87     return 0;
    88 }
    View Code
  • 相关阅读:
    记一次mqtt压测过程
    记项目过程中代码分支管理
    测试流程
    Docker与K8s的区别
    Mysql之pymysql
    Mysql常用简介
    JQuery
    CSS
    红外线接受程序 理解
    数码管流水灯升级程序理解
  • 原文地址:https://www.cnblogs.com/fu11211129/p/4905497.html
Copyright © 2011-2022 走看看