"DFS" Code Answer's

You're definitely familiar with the best coding language C++ that developers use to develop their projects and they get all their queries like "DFS" answered properly. Developers are finding an appropriate answer about DFS related to the C++ coding language. By visiting this online portal developers get answers concerning C++ codes question like DFS. Enter your desired code related query in the search bar and get every piece of information about C++ code related question on DFS. 

DFS in c++

By VeNOMVeNOM on Dec 08, 2020
#include <bits/stdc++.h>
using namespace std;
 

class Graph {
    int V; 
 
 
    list<int>* adj;
 
  
    void DFSUtil(int v, bool visited[]);
 
public:
    Graph(int V);
 
    void addEdge(int v, int w);
 
  
    void DFS(int v);
};
 
Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V];
}
 
void Graph::addEdge(int v, int w)
{
    adj[v].push_back(w); 
}
 
void Graph::DFSUtil(int v, bool visited[])
{
   
    visited[v] = true;
    cout << v << " ";
 
   
    list<int>::iterator i;
    for (i = adj[v].begin(); i != adj[v].end(); ++i)
        if (!visited[*i])
            DFSUtil(*i, visited);
}
 

void Graph::DFS(int v)
{
   
    bool* visited = new bool[V];
    for (int i = 0; i < V; i++)
        visited[i] = false;
 
 
    DFSUtil(v, visited);
}
 

int main()
{
  
    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);
 
    cout << "Following is Depth First Traversal"
            " (starting from vertex 2) \n";
    g.DFS(2);
 
    return 0;
}

Add Comment

2

dfs python

By actuallyitsnathanielactuallyitsnathaniel on Oct 05, 2020
###############
#The Algorithm (In English):

# 1) Pick any node. 
# 2) If it is unvisited, mark it as visited and recur on all its 
#    adjacent nodes. 
# 3) Repeat until all the nodes are visited, or the node to be 
#    searched is found.


# The graph below (declared as a Python dictionary)
# is from the linked website and is used for the sake of
# testing the algorithm. Obviously, you will have your own
# graph to iterate through.
graph = {
    'A' : ['B','C'],
    'B' : ['D', 'E'],
    'C' : ['F'],
    'D' : [],
    'E' : ['F'],
    'F' : []
}

visited = set() # Set to keep track of visited nodes.


##################
# The Algorithm (In Code)

def dfs(visited, graph, node):
    if node not in visited:
        print (node)
        visited.add(node)
        for neighbour in graph[node]:
            dfs(visited, graph, neighbour)
            
# Driver Code to test in python yourself.
# Note that when calling this, you need to
# call the starting node. In this case it is 'A'.
dfs(visited, graph, 'A')

# NOTE: There are a few ways to do DFS, depending on what your
# variables are and/or what you want returned. This specific
# example is the most fleshed-out, yet still understandable,
# explanation I could find.

Source: www.educative.io

Add Comment

6

DFS explained

By Annoying AlpacaAnnoying Alpaca on Mar 19, 2021
 def depth_first_search(graph):
    visited, stack = set(), [root]
    while stack:
        vertex = stack.pop()
        if vertex not in visited:
            visited.add(vertex)
            stack.extend(graph[vertex] - visited)
    return visited

Source: brilliant.org

Add Comment

0

depth first search

By RangerRanger on Oct 11, 2020
# HAVE USED ADJACENY LIST
class Graph:
    def __init__(self,lst=None):
        self.lst=dict()
        if lst is None:
            pass
        else:
            self.lst=lst
    def find_path(self,start,end):
        self.checklist={}
        for i in self.lst.keys():
            self.checklist[i]=False
        self.checklist[start]=True
        store,extra=(self.explore(start,end))
        if store==False:
            print('No Path Found')
        else:
            print(extra)
    def explore(self,start,end):
        while True:
            q=[]        
            #print(self.checklist,q)
            q.append(start)
            flag=False            
            for i in self.lst[start]:
                if i==end:
                    q.append(i)
                    return True,q
                if self.checklist[i]:
                    pass
                else:
                    flag=True
                    self.checklist[i]=True
                    q.append(i)
                    break   
            if flag:
                store,extra=self.explore(q[-1],end) 
                if store==False:
                    q.pop()
                    if len(q)==0:return False
                    return self.explore(q[-1],end)
                elif store==None:
                    pass
                elif store==True:
                    q.pop()
                    q.extend(extra)
                    return True,q
            else:
                return False,None
    def __str__(self):return str(self.lst)
if __name__=='__main__':
    store={1: [2, 3, 4], 2: [3, 1], 3: [2, 1], 4: [5, 8, 1], 5: [4, 6, 7], 6: [5, 7, 9, 8], 7: [5, 6], 8: [4, 6, 9], 9: [6, 8, 10], 10: [9],11:[12,13]}
    a=Graph(store)
    a.find_path(1,11) # No Path Found 
    a.find_path(1,6)# [1, 4, 5, 6]    
    a.find_path(3,10)   # [3, 2, 1, 4, 5, 6, 9, 10] 
    a.find_path(4,10)# [4, 5, 6, 9, 10]
    print(a) #

Add Comment

0

DFS

By Itchy ImpalaItchy Impala on May 31, 2021
import java.io.*;
import java.util.*;
 

class Graph {
    private int V;                              //number of nodes
 
    private LinkedList<Integer> adj[];              //adjacency list
 
    public Graph(int v)
    {
        V = v;
        adj = new LinkedList[v];
        for (int i = 0; i < v; ++i)
{
          adj[i] = new LinkedList();
    	}
 
    void addEdge(int v, int w)
    {
        adj[v].add(w);                              //adding an edge to the adjacency list (edges are bidirectional in this example)
    }
 

    void DFSUtil(int vertex, boolean nodes[])
    {

        nodes[vertex] = true;                         //mark the node as explored
        System.out.print(vertex + " ");
        int a = 0;
 
        for (int i = 0; i < adj[vertex].size(); i++)  //iterate through the linked list and then propagate to the next few nodes
            {
                a = adj[vertex].get(i);
                if (!nodes[a])                    //only propagate to next nodes which haven't been explored
                {
                    DFSUtil(a, nodes);
                }
            }  
    }

    void DFS(int v)
    {
        boolean already[] = new boolean[V];             //initialize a new boolean array to store the details of explored nodes
        DFSUtil(v, already);
    }
 
    public static void main(String args[])
    {
        Graph g = new Graph(6);
 
        g.addEdge(0, 1);
        g.addEdge(0, 2);
        g.addEdge(1, 0);
        g.addEdge(1, 3);
        g.addEdge(2, 0);
        g.addEdge(2, 3);
        g.addEdge(3, 4);
        g.addEdge(3, 5);
        g.addEdge(4, 3);
        g.addEdge(5, 3);
 
        System.out.println(
            "Following is Depth First Traversal: ");
 
        g.DFS(0);
    }
}

Source: favtutor.com

Add Comment

1

DFS

By Annoyed AntelopeAnnoyed Antelope on Apr 28, 2021
print("found ans")

Source: purdue.brightspace.com

Add Comment

-2

All those coders who are working on the C++ based application and are stuck on DFS can get a collection of related answers to their query. Programmers need to enter their query on DFS related to C++ code and they'll get their ambiguities clear immediately. On our webpage, there are tutorials about DFS for the programmers working on C++ code while coding their module. Coders are also allowed to rectify already present answers of DFS while working on the C++ language code. Developers can add up suggestions if they deem fit any other answer relating to "DFS". Visit this developer's friendly online web community, CodeProZone, and get your queries like DFS resolved professionally and stay updated to the latest C++ updates. 

C++ answers related to "DFS"

View All C++ queries

C++ queries related to "DFS"

Browse Other Code Languages

CodeProZone