In this assignment, you will be asked to do two things related to graphs over a really cool data set. What I have done is to collect the city-to-city destinations for Spirit Airlines, the 7th largest airline in the United States. This airline services 58 airports in the United States, Canada, the Caribbean, Central America, and South America. There are 438 flights between distinct locations. Your job is to figure out the most efficient flight plan to get from a city of your choice that Spirit Airlines serves to some other city that it serves. We will do that in two different contexts: the fewest planes (a breadth first search) and the least time (Dijkstra’s algorithm).

You have been provided with a text file that includes the origin of a flight, the destination of a flight, and the amount of time that it takes (in minutes) to get from one location to another. You will need to construct a symbol table to do this.
Rather than to write algorithms to do this, search for implementations of BFS and Dijkstra for the language of your choice and then use that to complete this assignment. You will need to modify the algorithms slightly such that you can generate a text file of the results.

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from collections import deque

class node:
    def __init__(self, name): = name
       self.adj = [] # neighbor node
       self.prev = None
       self.visited = False

class graph:

    def __init__(self, file):
       self.nodes = []

    def setNode(self, file):
       i = 0
       with open(file) as fin:
            for line in fin:
                if i == 0:
                   ssplit = line.split()
                   name = ssplit[0]
                   source = None
                   for n in self.nodes:
                        if == name:
                            source = n
                   if source == None:
                        source = node(name)

                   name = ssplit[1]
                   dest = None
                   for n in self.nodes:
                        if == name:
                            dest = n
                   if dest == None:
                        dest = node(name)

                i = i + 1

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