main.py
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import math
from collections import defaultdict

graph = {
'S': ['A'],
'A': ['B'],
'B': ['C', 'E'],
'C': ['A', 'D'],
'D': [],
'E': []
}

vertex_to_idx = {
'S': 0,
'A': 1,
'B': 2,
'C': 3,
'D': 4,
'E': 5
}

weights = {
'S': {'A': 6},
'A': {'B': 3},
'B': {'C': 4, 'E': 5},
'C': {'A': -3, 'D': 3},
'D': {},
'E': {}
}

num_edges = 0
num_vertices = 0
reversed_graph = defaultdict(list)
for vertex, edges in graph.items():
num_edges += len(edges)
num_vertices += 1
for edge in edges:
# Reversing the graph so we can later find which
# the vertices directly lead to a particular vertex
reversed_graph[edge].append(vertex)

d = [[math.inf for x in range(num_vertices)] for y in range(num_edges + 1)]
d[0][vertex_to_idx['S']] = 0

for i in range(1, num_edges):
for z in graph.keys():
z_idx = vertex_to_idx[z]

# Initialize the shortest path to z to the
# path found in the previous subproblem.
# Only update if new paths are shorter
d[i][z_idx] = d[i-1][z_idx]

# The reversed graph lets us find which vertexes
for y in reversed_graph[z]:
y_idx = vertex_to_idx[y]
if d[i][z_idx] > (d[i-1][y_idx] + weights[y][z]):
d[i][z_idx] = d[i-1][y_idx] + weights[y][z]

print()