Python:实现O( E + V ） 中找到 0-1-graph 中的最短路径算法
from __future__ import annotations
from collections import deque
from collections.abc import Iterator
from dataclasses import dataclass
@dataclass
class Edge:
"""Weighted directed graph edge."""
destination_vertex: int
weight: int
class AdjacencyList:
"""Graph adjacency list."""
def __init__(self, size: int):
self._graph: list[list[Edge]] = [[] for _ in range(size)]
self._size = size
def __getitem__(self, vertex: int) -> Iterator[Edge]:
"""Get all the vertices adjacent to the given one."""
return iter(self._graph[vertex])
@property
def size(self):
return self._size
def add_edge(self, from_vertex: int, to_vertex: int, weight: int):
if weight not in (0, 1):
raise ValueError("Edge weight must be either 0 or 1.")
if to_vertex < 0 or to_vertex >= self.size:
raise ValueError("Vertex indexes must be in [0; size).")
self._graph[from_vertex].append(Edge(to_vertex, weight))
def get_shortest_path(self, start_vertex: int, finish_vertex: int) -> int | None:
queue = deque([start_vertex])
distances: list[int | None] = [None] * self.size
distances[start_vertex] = 0
while queue:
current_vertex = queue.popleft()
current_distance = distances[current_vertex]
if current_distance is None:
continue
for edge in self[current_vertex]:
new_distance = current_distance + edge.weight
dest_vertex_distance = distances[edge.destination_vertex]
if (
isinstance(dest_vertex_distance, int)
and new_distance >= dest_vertex_distance
):
continue
distances[edge.destination_vertex] = new_distance
if edge.weight == 0:
queue.appendleft(edge.destination_vertex)
else:
queue.append(edge.destination_vertex)
if distances[finish_vertex] is None:
raise ValueError("No path from start_vertex to finish_vertex.")
return distances[finish_vertex]
if __name__ == "__main__":
import doctest
doctest.testmod()
