Dijkstra.py

# 迪杰斯特拉算法

# 第一个散列表，记录边的权重
graph = {}
graph["start"] = {}
graph["start"]["a"] = 6
graph["start"]["b"] = 2

# print(graph["start"].keys())  # 获取起点的所有邻居
# print(graph["start"]["a"])  # 起点到A的权重
# print(graph["start"]["b"])  # 起点到B的权重

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

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

graph["fin"] = {}  # 终点没有任何邻居

# 第二个散列表，存储每个节点的开销
infinity = float("inf")
costs = {}
costs["a"] = 6
costs["b"] = 2
costs["fin"] = infinity

# 第三个散列表，存储父节点
parents = {}
parents["a"] = "start"
parents["b"] = "start"
parents["fin"] = None

# 建立数组，记录处理过的节点
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while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


if __name__ == '__main__':
    print("Cost from the start to each node:")
    print(costs)
    # print(graph)
    # print(parents)