Python Dijkstra算法

Question

我正在尝试编写Dijkstra的算法，但是我在如何在代码中“说”某些事情上举步维艰。 要可视化，下面是我希望使用数组表示的列：

   max_nodes  
   A  B  C         Length       Predecessor       Visited/Unvisited
A 0  1   2             -1                                              U
B 1  0   1             -1                                              U
C 2  1   0             -1                                              U

因此，将有几个数组，如下面的代码所示：

def dijkstra (graph, start, end)

network[max_nodes][max_nodes]
state  [max_nodes][length]
state2 [max_nodes][predecessor]
state3 [max_nodes][visited]
initialNode = 0

    for nodes in graph:
      D[max_nodes][length] = -1
      P[max_nodes][predecessor] = ""
      V[max_nodes][visited] = false

      for l in graph:

       length = lengthFromSource[node] + graph[node][l]
          if length < lengthFromSourceNode[w]:
             state[l][length] = x
             state2[l][predecessor] 
             state3[l][visited] = true
          x +=1

粗体部分是我一直坚持的地方-我正在尝试实现算法的这一部分：

3。对于当前节点，考虑其所有未访问的邻居并计算它们的暂定距离。例如，如果当前节点（A）的距离为6，并且将其与另一节点（B）连接的边为2，则通过A到B的距离将为6+2=8。如果此距离小于先前记录的距离，请覆盖该距离
四。当我们考虑完当前节点的所有邻居时，将其标记为已访问。已访问的节点将不再被检查；它现在记录的距离是最终的和最小的

我想我是在正确的轨道上，我只是停留在如何说'从一个节点开始，得到从源到一个节点的长度，如果长度较小，覆盖上一个值，然后移动到下一个节点

Answer 1

首先，我假设这是一个家庭作业问题，最好的建议是不要费心自己写，而是在web上找到一个现有的实现。Here's one that looks pretty good, for example。

假设您需要重新设计轮子，这里引用的代码使用字典来存储节点数据。所以你给它喂食类似于：

{ 
  's': {'u' : 10, 'x' : 5}, 
  'u': {'v' : 1, 'x' : 2}, 
  'v': {'y' : 4}, 
  'x': {'u' : 3, 'v' : 9, 'y' : 2}, 
  'y': {'s' : 7, 'v' : 6}
}

这似乎是一种更直观的表示图形信息的方法。访问的节点和距离也可以保存在字典中。

Answer 2

我还用字典来存储网络。

创建网络词典（用户提供）

net = {'0':{'1':100, '2':300},
       '1':{'3':500, '4':500, '5':100},
       '2':{'4':100, '5':100},
       '3':{'5':20},
       '4':{'5':20},
       '5':{}
       }

最短路径算法（用户需要指定起点和终点节点）

def dijkstra(net, s, t):
    # sanity check
    if s == t:
        return "The start and terminal nodes are the same. Minimum distance is 0."
    if net.has_key(s)==False:
        return "There is no start node called " + str(s) + "."
    if net.has_key(t)==False:
        return "There is no terminal node called " + str(t) + "."
    # create a labels dictionary
    labels={}
    # record whether a label was updated
    order={}
    # populate an initial labels dictionary
    for i in net.keys():
        if i == s: labels[i] = 0 # shortest distance form s to s is 0
        else: labels[i] = float("inf") # initial labels are infinity
    from copy import copy
    drop1 = copy(labels) # used for looping
    ## begin algorithm
    while len(drop1) > 0:
        # find the key with the lowest label
        minNode = min(drop1, key = drop1.get) #minNode is the node with the smallest label
        # update labels for nodes that are connected to minNode
        for i in net[minNode]:
            if labels[i] > (labels[minNode] + net[minNode][i]):
                labels[i] = labels[minNode] + net[minNode][i]
                drop1[i] = labels[minNode] + net[minNode][i]
                order[i] = minNode
        del drop1[minNode] # once a node has been visited, it's excluded from drop1
    ## end algorithm
    # print shortest path
    temp = copy(t)
    rpath = []
    path = []
    while 1:
        rpath.append(temp)
        if order.has_key(temp): temp = order[temp]
        else: return "There is no path from " + str(s) + " to " + str(t) + "."
        if temp == s:
            rpath.append(temp)
            break
    for j in range(len(rpath)-1,-1,-1):
        path.append(rpath[j])
    return "The shortest path from " + s + " to " + t + " is " + str(path) + ". Minimum distance is " + str(labels[t]) + "."

# Given a large random network find the shortest path from '0' to '5'
print dijkstra(net=randNet(), s='0', t='5')