In previous , BFS only with a particular vertex is performed i.e. it is assumed that all vertices are reachable from the starting vertex. But in the case of disconnected graph or any vertex that is unreachable from all vertex, the previous implementation will not give the desired output, so in this post, a modification is done in BFS.
All vertices are reachable. So, for above graph simple will work
As in above graph a vertex 1 is unreachable from all vertex, so simple BFS wouldn’t work for it.
# Python3 implementation of modified BFS
import queue
# A utility function to add an edge
# in an undirected graph.
def addEdge(adj, u, v):
adj[u].append(v)
# A utility function to do BFS of
# graph from a given vertex u.
def BFSUtil(u, adj, visited):
# Create a queue for BFS
q = queue.Queue()
# Mark the current node as visited
# and enqueue it
visited[u] = True
q.put(u)
# 'i' will be used to get all adjacent
# vertices 4 of a vertex list<int>::iterator i
while(not q.empty()):
# Dequeue a vertex from queue
# and print it
u = q.queue[0]
print(u, end = " ")
q.get()
# Get all adjacent vertices of the
# dequeued vertex s. If an adjacent
# has not been visited, then mark
# it visited and enqueue it
i = 0
while i != len(adj[u]):
if (not visited[adj[u][i]]):
visited[adj[u][i]] = True
q.put(adj[u][i])
i += 1
# This function does BFSUtil() for all
# unvisited vertices.
def BFS(adj, V):
visited = [False] * V
for u in range(V):
if (visited[u] == False):
BFSUtil(u, adj, visited)
# Driver code
if __name__ == '__main__':
V = 5
adj = [[] for i in range(V)]
addEdge(adj, 0, 4)
addEdge(adj, 1, 2)
addEdge(adj, 1, 3)
addEdge(adj, 1, 4)
addEdge(adj, 2, 3)
addEdge(adj, 3, 4)
BFS(adj, V)
# This code is contributed by PranchalK
