代价树搜索策略求解交通图中两地之间的最短路径python实验代码

以下是一个基于代价树搜索策略的最短路径实验代码示例，可以用于求解交通图中两地之间的最短路径问题：

import heapq

def dijkstra(graph, start, end):
    # 初始化距离字典和前置节点字典
    dist = {node: float('inf') for node in graph}
    dist[start] = 0
    previous = {node: None for node in graph}

    # 初始化代价树
    cost_tree = {node: {} for node in graph}
    for neighbor, weight in graph[start].items():
        cost_tree[neighbor][start] = weight

    # 开始搜索
    heap = [(0, start)]
    visited = set()
    while heap:
        node_dist, node = heapq.heappop(heap)
        if node == end:
            break
        if node in visited:
            continue
        visited.add(node)
        for neighbor, weight in graph[node].items():
            if neighbor in visited:
                continue
            cost = node_dist + weight
            if cost < dist[neighbor]:
                dist[neighbor] = cost
                previous[neighbor] = node
                cost_tree[neighbor] = {node: weight}
                heapq.heappush(heap, (cost, neighbor))
            elif cost == dist[neighbor]:
                cost_tree[neighbor][node] = weight

    # 回溯路径
    path = []
    node = end
    while node is not None:
        path.append(node)
        node = previous[node]
    path.reverse()

    return path, dist[end]

# 测试代码
graph = {
    'A': {'B': 5, 'C': 1},
    'B': {'A': 5, 'C': 2, 'D': 1},
    'C': {'A': 1, 'B': 2, 'D': 4, 'E': 8},
    'D': {'B': 1, 'C': 4, 'E': 3, 'F': 6},
    'E': {'C': 8, 'D': 3},
    'F': {'D': 6}
}
start = 'A'
end = 'F'
path, dist = dijkstra(graph, start, end)
print("最短路径：", path)
print("最短距离：", dist)

该代码使用了代价树搜索策略来求解最短路径。其中，`graph`变量表示图的邻接表，`start`和`end`变量分别表示起点和终点。该算法会返回一条最短路径以及对应的距离。