“Belman-ford” Réponses codées

Belman-ford

# Python3 program for Bellman-Ford's single source
# shortest path algorithm.
 
# Class to represent a graph
class Graph:
 
    def __init__(self, vertices):
        self.V = vertices # No. of vertices
        self.graph = []
 
    # function to add an edge to graph
    def addEdge(self, u, v, w):
        self.graph.append([u, v, w])
         
    # utility function used to print the solution
    def printArr(self, dist):
        print("Vertex Distance from Source")
        for i in range(self.V):
            print("{0}\t\t{1}".format(i, dist[i]))
     
    # The main function that finds shortest distances from src to
    # all other vertices using Bellman-Ford algorithm. The function
    # also detects negative weight cycle
    def BellmanFord(self, src):
 
        # Step 1: Initialize distances from src to all other vertices
        # as INFINITE
        dist = [float("Inf")] * self.V
        dist[src] = 0
 
 
        # Step 2: Relax all edges |V| - 1 times. A simple shortest
        # path from src to any other vertex can have at-most |V| - 1
        # edges
        for _ in range(self.V - 1):
            # Update dist value and parent index of the adjacent vertices of
            # the picked vertex. Consider only those vertices which are still in
            # queue
            for u, v, w in self.graph:
                if dist[u] != float("Inf") and dist[u] + w < dist[v]:
                        dist[v] = dist[u] + w
 
        # Step 3: check for negative-weight cycles. The above step
        # guarantees shortest distances if graph doesn't contain
        # negative weight cycle. If we get a shorter path, then there
        # is a cycle.
 
        for u, v, w in self.graph:
                if dist[u] != float("Inf") and dist[u] + w < dist[v]:
                        print("Graph contains negative weight cycle")
                        return
                         
        # print all distance
        self.printArr(dist)
 
g = Graph(5)
g.addEdge(0, 1, -1)
g.addEdge(0, 2, 4)
g.addEdge(1, 2, 3)
g.addEdge(1, 3, 2)
g.addEdge(1, 4, 2)
g.addEdge(3, 2, 5)
g.addEdge(3, 1, 1)
g.addEdge(4, 3, -3)
 
# Print the solution
g.BellmanFord(0)
 
# Initially, Contributed by Neelam Yadav
# Later On, Edited by Himanshu Garg