迪克斯特拉(总权重最小路径)

最新推荐文章于 2024-08-13 16:25:55 发布

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通过迪克斯特拉算法，逐步找到图中从起点到每个节点的最短路径，不断更新邻居节点的开销，直到遍历所有节点，从而计算得出全局的最短路径。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1.找出"最便宜"的节点,即可在最短时间内到达的节点
2.更新该节点的邻居和开销,检查是否有前往他们的更短路劲,如果有,就更新其开销
3.重复这个过程,直到对图中的每个节点都这样做
4.计算最短路径

graph = {}
graph["start"] = {}
graph["start"]["a"] = 6
graph["start"]["b"] = 2
# print(graph["start"].keys())
# print(graph["start"]["a"])

graph["a"] = {}
graph["a"]["fin"] = 1

graph["b"] = {}
graph["b"]["a"] = 3
graph["b"]["fin"] = 5
graph["fin"] = {}

infinity = float("inf")
costs = {}
costs["a"] = 6
costs["b"] = 2
costs["fin"] = infinity

parents = {}
parents["a"] = "start"
parents["b"] = "start"
parents["fin"] = None

processed = []


def find_lowest_cost_node(costs):
    lowest_cost = float("inf")
    lowest_cost_node = None
    for node in costs:  # 遍历所有的节点
        cost = costs[node]
        if cost < lowest_cost and node not in processed:  # 如果当前节点的开销更低且尚未处理过
            lowest_cost = cost  # 就将其视为开销最低的节点
            lowest_cost_node = node
    return lowest_cost_node


node = find_lowest_cost_node(costs)  # 在未处理的节点中找出开销最小的节点
while node is not None:  # 这个循环在所有节点都被处理过后结束
    cost = costs[node]
    neighbors = graph[node]
    for n in neighbors.keys():  # 遍历当前节点的所有邻居
        new_cost = cost + neighbors[n]
        if costs[n] > new_cost:  # 如果经当前节点前往该邻居更近
            costs[n] = new_cost  # 就更新该邻居的开销
            parents[n] = node  # 同时将该邻居的父节点设置为当前节点
    processed.append(node)  # 将当前节点标记为处理过
    node = find_lowest_cost_node(costs)  # 找出接下来要处理的节点,并循环
print(graph)
print(costs)
print(parents)