求解起点和终点不确定的带权图的最短路径的代码

这是一个比较复杂的问题，需要使用一些图论算法。以下是一种基于Dijkstra算法的实现，可以求解起点和终点不确定的带权图的最短路径，你可以参考一下。

首先，我们需要定义一个图的数据结构。可以使用邻接表来表示图。

class Graph:
    def __init__(self, vertices):
        self.V = vertices
        self.graph = defaultdict(list)

    def add_edge(self, u, v, w):
        self.graph[u].append((v, w))
        self.graph[v].append((u, w))

接下来，我们实现Dijkstra算法来求解最短路径。

import heapq

def dijkstra(graph, source): 
    distance = [float("inf")]*graph.V
    distance[source] = 0
    heap = [(0, source)]
  
    while heap:
        (dist, current) = heapq.heappop(heap)
        if dist > distance[current]:
            continue
        for neighbor, weight in graph.graph[current]:
            w = weight + dist
            if w < distance[neighbor]:
                distance[neighbor] = w
                heapq.heappush(heap, (w, neighbor))
  
    return distance

最后，我们可以将这两个函数组合起来，来计算任意两个节点的最短距离。以下是完整的代码。

from collections import defaultdict
import heapq

class Graph:
    def __init__(self, vertices):
        self.V = vertices
        self.graph = defaultdict(list)
  
    def add_edge(self, u, v, w):
        self.graph[u].append((v, w))
        self.graph[v].append((u, w))
  
def dijkstra(graph, source): 
    distance = [float("inf")]*graph.V
    distance[source] = 0
    heap = [(0, source)]
  
    while heap:
        (dist, current) = heapq.heappop(heap)
        if dist > distance[current]:
            continue
        for neighbor, weight in graph.graph[current]:
            w = weight + dist
            if w < distance[neighbor]:
                distance[neighbor] = w
                heapq.heappush(heap, (w, neighbor))
  
    return distance
  
def shortest_path(graph):
    result = []
  
    for i in range(graph.V):
        dist = dijkstra(graph, i)
        result.append(dist)
  
    return result

注意，这只是其中一种实现方式，并不是唯一的解决方案。你可以根据需求选择适合的算法。