中文字幕av专区_日韩电影在线播放_精品国产精品久久一区免费式_av在线免费观看网站

溫馨提示×

溫馨提示×

您好,登錄后才能下訂單哦!

密碼登錄×
登錄注冊×
其他方式登錄
點擊 登錄注冊 即表示同意《億速云用戶服務條款》

python中怎么實現一個Dijkstra算法

發布時間:2021-07-10 11:39:41 來源:億速云 閱讀:327 作者:Leah 欄目:大數據

python中怎么實現一個Dijkstra算法,針對這個問題,這篇文章詳細介紹了相對應的分析和解答,希望可以幫助更多想解決這個問題的小伙伴找到更簡單易行的方法。


"""

Dijkstra algorithm

graphdict={"A":[("B",6),("C",3)], "B":[("C",2),("D",5)],"C":[("B",2),("D",3),("E",4)],\

         "D":[("B",5),("C",3),("E",2),("F",3)],"E":[("C",4),("D",2),("F",5)],"F":[("D",3),"(E",5)]})

assert: start node must be zero in-degree

"""


def Dijkstra(startNode, endNode, graphdict=None):

    S=[startNode]

    V=[]

    for node in graphdict.keys():

        if node !=startNode:

            V.append(node)

    #distance dict from startNode

    dist={}

    for node in V:

        dist[node]=float('Inf')


    while len(V)>0:

        center = S[-1] # get final node for S as the new center node

        minval = ("None",float("Inf"))

        for node,d in graphdict[center]:

            if node not in V:

                continue

            #following is the key logic.If S length is bigger than 1,need to get the final ele of S, which is the center point in current

            #iterator, and distance between start node and center node is startToCenterDist; d is distance between node

            # among out-degree for center point; dist[node] is previous distance to start node, possibly Inf or a updated value

            # so if startToCenterDist+d is less than dist[node], then it shows we find a shorter distance.

            if len(S)==1:

                dist[node] = d

            else:

                startToCenterDist = dist[center]

                if startToCenterDist + d < dist[node]:

                    dist[node] = startToCenterDist + d

            #this is the method to find a new center node and

            # it's the minimum distance among out-degree nodes for center node

            if d < minval[1]:

                minval = (node,d)

        V.remove(minval[0])

        S.append(minval[0]) # append node with min val

    return dist


03

測試

python中怎么實現一個Dijkstra算法

求出以上圖中,從A到各個節點的最短路徑:

shortestRoad = Dijkstra("A","F",graphdict={"A":[("B",6),("C",3)], "B":[("C",2),("D",5)],\

                            "C":[("B",2),("D",3),("E",4)],\

                            "D":[("B",5),("C",3),("E",2),("F",3)],\

                            "E":[("C",4),("D",2),("F",5)],"F":[("D",3),("E",5)]})

mystr = "shortest distance from A begins to "

for key,shortest in shortestRoad.items():

    print(mystr+ str(key) +" is: " + str(shortest) )

打印結果如下:

shortest distance from A begins to B is: 5

shortest distance from A begins to C is: 3

shortest distance from A begins to D is: 6

shortest distance from A begins to E is: 7

shortest distance from A begins to F is: 9

關于python中怎么實現一個Dijkstra算法問題的解答就分享到這里了,希望以上內容可以對大家有一定的幫助,如果你還有很多疑惑沒有解開,可以關注億速云行業資訊頻道了解更多相關知識。

向AI問一下細節

免責聲明:本站發布的內容(圖片、視頻和文字)以原創、轉載和分享為主,文章觀點不代表本網站立場,如果涉及侵權請聯系站長郵箱:is@yisu.com進行舉報,并提供相關證據,一經查實,將立刻刪除涉嫌侵權內容。

AI

云南省| 杭州市| 右玉县| 海南省| 富民县| 安化县| 泰和县| 进贤县| 突泉县| 阜康市| 东莞市| 泰州市| 马边| 山东省| 阿勒泰市| 夏邑县| 临颍县| 库尔勒市| 卢龙县| 淳安县| 正阳县| 奉节县| 垫江县| 原阳县| 兰考县| 保德县| 银川市| 沾益县| 会昌县| 朔州市| 蓬安县| 民县| 临泽县| 宜宾县| 扬州市| 呼伦贝尔市| 临桂县| 邯郸县| 浦县| 澄江县| 台湾省|