The Flight Plan - Hackerearth Problem Solution Using Python

Here , To solve this problem we have use the BFS algorithm using python in which we just add a "dist" list in which we store distance of every node from its parent node. And we have use dictionary named "dd" in which we store its parent node if we print that dictionary we get the path for for each node from source given node.

For the Problem goto below link and search for problem: The Flight Plan

Code :

from collections import defaultdict class Graph: def __init__(self): self.graph = defaultdict(list) def addEdge(self,u,v): self.graph[u].append(v) def BFS(self, s,l): dist=[0]*100 dd=defaultdict(list) dd[s].append(0) visited = [False] * 100#(len(self.graph)+1) queue = [] queue.append(s) visited[s] = True while queue: s = queue.pop(0) for i in self.graph[s]: if visited[i] == False: dist[i]=dist[s]+1 dd[i].append(s) for j in dd[s]: dd[i].append(j) queue.append(i) visited[i] = True print(dist[l]+1) return dd[l] def main(): l=list(map(int,input().split())) g=Graph() for i in range(l[1]): l2=list(map(int,input().split())) g.addEdge(l2[0],l2[1]) g.addEdge(l2[1],l2[0]) l3=list(map(int,input().split())) ans=g.BFS(l3[0],l3[1]) ans.reverse() for i in ans: if i==0: continue print(i,end=' ') print(l3[1],end=' ') main()

Social Networking Graph - Hackerearth Problem Solution Using Python

Here , To solve this problem we have use the BFS algorithm using python in which we just add a "dist" list in which we store distance of every node from its parent node. And then we see count nodes which are at a given distance from given source node.

For the Problem goto below link and search for problem: Social Networking Graph

Code :

from collections import defaultdict from collections import defaultdict class Graph: def __init__(self): self.graph = defaultdict(list) def addEdge(self,u,v): self.graph[u].append(v) def BFS(self, s,a): dist=[0]*100 visited = [False] * 100#(len(self.graph)+1) queue = [] queue.append(s) visited[s] = True while queue: s = queue.pop(0) for i in self.graph[s]: if visited[i] == False: dist[i]=dist[s]+1 #this is the dist list queue.append(i) visited[i] = True return dist.count(a) def main(): ml=[] l=list(map(int,input().split())) g=Graph() for i in range(l[1]): l2=list(map(int,input().split())) g.addEdge(l2[0],l2[1]) g.addEdge(l2[1],l2[0]) #graph is bidirectional n=int(input()) for j in range(n): ll=list(map(int,input().split())) ans=g.BFS(ll[0],ll[1]) ml.append(ans) for i in ml: print(i) main()

Monk and the Islands - Hackerearth Problem Solution Using Python

Here , To solve this problem we have use the BFS algorithm using python in which we just add a "dist" list in which we store distance of every node from its parent node.

For the Problem goto below link and search for problem: Monk and Islands

Code :

from collections import defaultdict from collections import defaultdict class Graph: def __init__(self): self.graph = defaultdict(list) def addEdge(self,u,v): self.graph[u].append(v) def BFS(self, s,l): dist=[0]*10000000 visited = [False] * 10000000#(len(self.graph)+1) queue = [] queue.append(s) visited[s] = True while queue: s = queue.pop(0) for i in self.graph[s]: if visited[i] == False: dist[i]=dist[s]+1 #this is the dist list queue.append(i) visited[i] = True return dist[l[0]] def main(): t=int(input()) ml=[] for i in range(t): l=list(map(int,input().split())) g=Graph() for i in range(l[1]): l2=list(map(int,input().split())) g.addEdge(l2[0],l2[1]) g.addEdge(l2[1],l2[0]) #graph is bidirectional ans=g.BFS(1,l) ml.append(ans) for i in ml: print(i) main()

Shortest Path Using Breadth First Search (BFS)

Shortest Path Using Breadth First Search (BFS)

Here , To find the shortest path using BFS first we have to know that the graph should always be unweighted graph .

If you know about the BFS algorithm then to find shortest path we have to add only one thing that is a dictionary named "dist" which records the distance of every node from given node As shown in below code.

Code :

from collections import defaultdict # This class represents a directed graph # using adjacency list representation class Graph: # Constructor def __init__(self): # default dictionary to store graph self.graph = defaultdict(list) # function to add an edge to graph def addEdge(self,u,v): self.graph[u].append(v) # Function to print a BFS of graph def BFS(self, s): dist=[0]*10 # Mark all the vertices as not visited visited = [False] * (len(self.graph)) # 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 and print it s = queue.pop(0) print (s, end = " ") # 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 in self.graph[s]: if visited[i] == False: dist[i]=dist[s]+1 queue.append(i) visited[i] = True print(dist) # Driver code # Create a graph given in # the above diagram g = Graph() g.addEdge(0, 1) g.addEdge(1,0) g.addEdge(0, 3) g.addEdge(3,0) g.addEdge(1, 2) g.addEdge(2,1) g.addEdge(3, 7) g.addEdge(7,3) g.addEdge(3, 4) g.addEdge(4,3) g.addEdge(4, 7) g.addEdge(7,4) g.addEdge(4, 6) g.addEdge(6,4) g.addEdge(7, 6) g.addEdge(6,7) g.addEdge(6, 5) g.addEdge(5,6) print ("Following is Breadth First Traversal" " (starting from vertex 2)") g.BFS(2)