Encuentra si hay un camino entre dos vértices en un gráfico dirigido

Dado un gráfico dirigido y dos vértices en él, compruebe si hay un camino desde el primer vértice dado hasta el segundo. 
Ejemplo: 

Consider the following Graph:


Input : (u, v) = (1, 3)
Output: Yes
Explanation: There is a path from 1 to 3, 1 -> 2 -> 3

Input : (u, v) = (3, 6)
Output: No
Explanation: There is no path from 3 to 6

Enfoque: Se puede utilizar la búsqueda primero en amplitud (BFS) o la búsqueda primero en profundidad (DFS) para encontrar la ruta entre dos vértices. Tome el primer vértice como fuente en BFS (o DFS), siga el BFS estándar (o DFS). Si el segundo vértice se encuentra en nuestro recorrido, devuelve verdadero; de lo contrario, devuelve falso.

Algoritmo BFS: 

1. La implementación a continuación utiliza BFS.
2. Cree una cola y una array visitada inicialmente rellena con 0, de tamaño V, donde V es un número de vértices.
3. Inserte el Node inicial en la cola, es decir, presione u en la cola y márquelo como visitado.
4. Ejecute un bucle hasta que la cola no esté vacía.
5. Dequeue el elemento frontal de la cola. Iterar todos sus elementos adyacentes. Si alguno de los elementos adyacentes es el destino, devuelve verdadero. Empuje todos los vértices adyacentes y no visitados en la cola y márquelos como visitados.
6. Devuelve false ya que no se alcanza el destino en BFS.

Implementación: códigos C++, Java y Python que usan BFS para encontrar la accesibilidad del segundo vértice desde el primer vértice. 

// C++ program to check if there is exist a path between two vertices
// of a graph.
#include<iostream>
#include <list>
using namespace std;
 
// This class represents a directed graph using adjacency list
// representation
class Graph
{
    int V;    // No. of vertices
    list<int> *adj;    // Pointer to an array containing adjacency lists
public:
    Graph(int V);  // Constructor
    void addEdge(int v, int w); // function to add an edge to graph
    bool isReachable(int s, int d); 
};
 
Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V];
}
 
void Graph::addEdge(int v, int w)
{
    adj[v].push_back(w); // Add w to v’s list.
}
 
// A BFS based function to check whether d is reachable from s.
bool Graph::isReachable(int s, int d)
{
    // Base case
    if (s == d)
      return true;
 
    // Mark all the vertices as not visited
    bool *visited = new bool[V];
    for (int i = 0; i < V; i++)
        visited[i] = false;
 
    // Create a queue for BFS
    list<int> queue;
 
    // Mark the current node as visited and enqueue it
    visited[s] = true;
    queue.push_back(s);
 
    // it will be used to get all adjacent vertices of a vertex
    list<int>::iterator i;
 
    while (!queue.empty())
    {
        // Dequeue a vertex from queue and print it
        s = queue.front();
        queue.pop_front();
 
        // Get all adjacent vertices of the dequeued vertex s
        // If a adjacent has not been visited, then mark it visited
        // and enqueue it
        for (i = adj[s].begin(); i != adj[s].end(); ++i)
        {
            // If this adjacent node is the destination node, then
            // return true
            if (*i == d)
                return true;
 
            // Else, continue to do BFS
            if (!visited[*i])
            {
                visited[*i] = true;
                queue.push_back(*i);
            }
        }
    }
     
    // If BFS is complete without visiting d
    return false;
}
 
// Driver program to test methods of graph class
int main()
{
    // Create a graph given in the above diagram
    Graph g(4);
    g.addEdge(0, 1);
    g.addEdge(0, 2);
    g.addEdge(1, 2);
    g.addEdge(2, 0);
    g.addEdge(2, 3);
    g.addEdge(3, 3);
 
    int u = 1, v = 3;
    if(g.isReachable(u, v))
        cout<< "\n There is a path from " << u << " to " << v;
    else
        cout<< "\n There is no path from " << u << " to " << v;
 
    u = 3, v = 1;
    if(g.isReachable(u, v))
        cout<< "\n There is a path from " << u << " to " << v;
    else
        cout<< "\n There is no path from " << u << " to " << v;
 
    return 0;
}

// Java program to check if there is exist a path between two vertices
// of a graph.
import java.io.*;
import java.util.*;
import java.util.LinkedList;
 
// This class represents a directed graph using adjacency list
// representation
class Graph
{
    private int V;   // No. of vertices
    private LinkedList<Integer> adj[]; //Adjacency List
 
    //Constructor
    Graph(int v)
    {
        V = v;
        adj = new LinkedList[v];
        for (int i=0; i<v; ++i)
            adj[i] = new LinkedList();
    }
 
    //Function to add an edge into the graph
    void addEdge(int v,int w)  {   adj[v].add(w);   }
 
    //prints BFS traversal from a given source s
    Boolean isReachable(int s, int d)
    {
        LinkedList<Integer>temp;
 
        // Mark all the vertices as not visited(By default set
        // as false)
        boolean visited[] = new boolean[V];
 
        // Create a queue for BFS
        LinkedList<Integer> queue = new LinkedList<Integer>();
 
        // Mark the current node as visited and enqueue it
        visited[s]=true;
        queue.add(s);
 
        // 'i' will be used to get all adjacent vertices of a vertex
        Iterator<Integer> i;
        while (queue.size()!=0)
        {
            // Dequeue a vertex from queue and print it
            s = queue.poll();
 
            int n;
            i = adj[s].listIterator();
 
            // Get all adjacent vertices of the dequeued vertex s
            // If a adjacent has not been visited, then mark it
            // visited and enqueue it
            while (i.hasNext())
            {
                n = i.next();
 
                // If this adjacent node is the destination node,
                // then return true
                if (n==d)
                    return true;
 
                // Else, continue to do BFS
                if (!visited[n])
                {
                    visited[n] = true;
                    queue.add(n);
                }
            }
        }
 
        // If BFS is complete without visited d
        return false;
    }
 
    // Driver method
    public static void main(String args[])
    {
        // Create a graph given in the above diagram
        Graph g = new Graph(4);
        g.addEdge(0, 1);
        g.addEdge(0, 2);
        g.addEdge(1, 2);
        g.addEdge(2, 0);
        g.addEdge(2, 3);
        g.addEdge(3, 3);
 
        int u = 1;
        int v = 3;
        if (g.isReachable(u, v))
            System.out.println("There is a path from " + u +" to " + v);
        else
            System.out.println("There is no path from " + u +" to " + v);;
 
        u = 3;
        v = 1;
        if (g.isReachable(u, v))
            System.out.println("There is a path from " + u +" to " + v);
        else
            System.out.println("There is no path from " + u +" to " + v);;
    }
}
// This code is contributed by Aakash Hasija

# program to check if there is exist a path between two vertices
# of a graph
 
from collections import defaultdict
  
#This class represents a directed graph using adjacency list representation
class Graph:
  
    def __init__(self,vertices):
        self.V= vertices #No. of vertices
        self.graph = defaultdict(list) # default dictionary to store graph
  
    # function to add an edge to graph
    def addEdge(self,u,v):
        self.graph[u].append(v)
      
     # Use BFS to check path between s and d
    def isReachable(self, s, d):
        # Mark all the vertices as not visited
        visited =[False]*(self.V)
  
        # Create a queue for BFS
        queue=[]
  
        # Mark the source node as visited and enqueue it
        queue.append(s)
        visited[s] = True
  
        while queue:
 
            #Dequeue a vertex from queue
            n = queue.pop(0)
             
            # If this adjacent node is the destination node,
            # then return true
            if n == d:
                   return True
 
            #  Else, continue to do BFS
            for i in self.graph[n]:
                if visited[i] == False:
                    queue.append(i)
                    visited[i] = True
         # If BFS is complete without visited d
        return False
  
# Create a graph given in the above diagram
g = Graph(4)
g.addEdge(0, 1)
g.addEdge(0, 2)
g.addEdge(1, 2)
g.addEdge(2, 0)
g.addEdge(2, 3)
g.addEdge(3, 3)
 
u =1; v = 3
 
if g.isReachable(u, v):
    print("There is a path from %d to %d" % (u,v))
else :
    print("There is no path from %d to %d" % (u,v))
 
u = 3; v = 1
if g.isReachable(u, v) :
    print("There is a path from %d to %d" % (u,v))
else :
    print("There is no path from %d to %d" % (u,v))
 
#This code is contributed by Neelam Yadav

// C# program to check if there is
// exist a path between two vertices
// of a graph.
using System;
using System.Collections;
using System.Collections.Generic;
 
// This class represents a directed
// graph using adjacency list
// representation
class Graph
{
  private int V; // No. of vertices
  private LinkedList<int>[] adj; //Adjacency List
 
  // Constructor
  Graph(int v)
  {
    V = v;
    adj = new LinkedList<int>[v];
    for (int i = 0; i < v; ++i)
      adj[i] = new LinkedList<int>();
  }
 
  // Function to add an edge into the graph
  void addEdge(int v, int w)
  {
    adj[v].AddLast(w);
  }
 
  // prints BFS traversal from a given source s
  bool isReachable(int s, int d)
  {
    // LinkedList<int> temp = new LinkedList<int>();
 
    // Mark all the vertices as not visited(By default set
    // as false)
    bool[] visited = new bool[V];
 
    // Create a queue for BFS
    LinkedList<int> queue = new LinkedList<int>();
 
    // Mark the current node as visited and enqueue it
    visited[s] = true;
    queue.AddLast(s);
 
    // 'i' will be used to get all adjacent vertices of a vertex
    IEnumerator i;     
    while (queue.Count != 0)
    {
 
      // Dequeue a vertex from queue and print it
      s = queue.First.Value;
      queue.RemoveFirst();
      int n;
      i = adj[s].GetEnumerator();
 
      // Get all adjacent vertices of the dequeued vertex s
      // If a adjacent has not been visited, then mark it
      // visited and enqueue it
      while (i.MoveNext())
      {
        n = (int)i.Current;
 
        // If this adjacent node is the destination node,
        // then return true
        if (n == d)
          return true;
 
        // Else, continue to do BFS
        if (!visited[n])
        {
          visited[n] = true;
          queue.AddLast(n);
        }
      }
    }
 
    // If BFS is complete without visited d
    return false;
  }
 
  // Driver method
  public static void Main(string[] args)
  {
 
    // Create a graph given in the above diagram
    Graph g = new Graph(4);
    g.addEdge(0, 1);
    g.addEdge(0, 2);
    g.addEdge(1, 2);
    g.addEdge(2, 0);
    g.addEdge(2, 3);
    g.addEdge(3, 3);
    int u = 1;
    int v = 3;
    if (g.isReachable(u, v))
      Console.WriteLine("There is a path from " + u + " to " + v);
    else
      Console.WriteLine("There is no path from " + u + " to " + v);
    u = 3;
    v = 1;
    if (g.isReachable(u, v))
      Console.WriteLine("There is a path from " + u + " to " + v);
    else
      Console.WriteLine("There is no path from " + u + " to " + v);
  }
}
 
// This code is contributed by sanjeev2552

<script>
// Javascript program to check if there is exist a path between two vertices
// of a graph.
let  V;
let adj;
 
function Graph( v)
{
        V = v;
        adj = new Array(v);
        for (let i = 0; i < v; ++i)
            adj[i] = [];
}
 
// Function to add an edge into the graph
function addEdge(v,w)
{
    adj[v].push(w);
}
 
// prints BFS traversal from a given source s
function isReachable(s,d)
{
    let temp;
  
        // Mark all the vertices as not visited(By default set
        // as false)
        let visited = new Array(V);
         for(let i = 0; i < V; i++)
            visited[i] = false;
             
        // Create a queue for BFS
        let queue = [];
  
        // Mark the current node as visited and enqueue it
        visited[s] = true;
        queue.push(s);
  
        while (queue.length != 0)
        {
            // Dequeue a vertex from queue and print it
            n = queue.shift();
              
            if(n == d)
                return true;
            for(let i = 0; i < adj[n].length; i++)
            {
                if (visited[adj[n][i]] == false)
                {
                    queue.push(adj[n][i]);
                    visited[adj[n][i]] = true;
                }
            }
             
        }
  
        // If BFS is complete without visited d
        return false;
}
 
// Driver method
Graph(4);
addEdge(0, 1);
addEdge(0, 2);
addEdge(1, 2);
addEdge(2, 0);
addEdge(2, 3);
addEdge(3, 3);
 
let u = 1;
let v = 3;
if (isReachable(u, v))
    document.write("There is a path from " + u +" to " + v+"<br>");
else
    document.write("There is no path from " + u +" to " + v+"<br>");
 
u = 3;
v = 1;
if (isReachable(u, v))
    document.write("There is a path from " + u +" to " + v+"<br>");
else
    document.write("There is no path from " + u +" to " + v+"<br>");
 
// This code is contributed by avanitrachhadiya2155
</script>

 There is a path from 1 to 3
 There is no path from 3 to 1

Análisis de Complejidad:  

• Complejidad de tiempo: O(V+E) donde V es el número de vértices en el gráfico y E es el número de aristas en el gráfico.
• Complejidad espacial: O(V). 
  Puede haber como máximo V elementos en la cola. Entonces el espacio necesario es O(V).

Algoritmo DFS:

1. if start==end devuelve 1 ya que tenemos que llegar a nuestro destino.
2. Marcar inicio como visitado .
3. Atraviese los vértices de inicio conectados directamente y repita la función dfs para cada vértice inexplorado.
4. devuelve 0 si no llegamos a nuestro destino.

Implementación :

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
 
vector<ll> adj[100000];
bool visited[100000];
 
bool dfs(int start, int end)
{
    if (start == end)
        return true;
    visited[start] = 1;
    for (auto x : adj[start]) {
        if (!visited[x])
            if (dfs(x, end))
                return true;
    }
    return false;
}
 
int main()
{
 
    int V = 4;
    vector<ll> members = { 2, 5, 7, 9 };
 
    int E = 4;
    vector<pair<ll, ll> > connections
        = { { 2, 9 }, { 7, 2 }, { 7, 9 }, { 9, 5 } };
 
    for (int i = 0; i < E; i++)
        adj[connections[i].first].push_back(
            connections[i].second);
 
    int sender = 7, receiver = 9;
 
    if (dfs(sender, receiver))
        cout << "1";
    else
        cout << "0";
 
    return 0;
}
// this code is contributed by prophet1999

1

Análisis de Complejidad:  

Complejidad de tiempo: O(V+E) donde V es el número de vértices en el gráfico y E es el número de aristas en el gráfico.
Complejidad espacial: O(V). 
Puede haber como máximo V elementos en la pila. Entonces el espacio necesario es O(V).

Compensaciones entre BFS y DFS: 

La búsqueda primero en amplitud puede ser útil para encontrar el camino más corto entre Nodes, y la búsqueda primero en profundidad puede atravesar un Node adyacente muy profundamente antes de llegar a los vecinos inmediatos. 
Como ejercicio, pruebe una versión extendida del problema donde también se necesita el camino completo entre dos vértices.