我正在尝试编写Dijkstra的算法,但我正在努力解决如何在代码中"说出"某些事情.为了可视化,这里是我想要使用数组表示的列:
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
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所以,会有几个数组,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
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粗体部分是我坚持的地方 - 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其暂定距离.例如,如果当前节点(A)的距离为6,并且将其与另一个节点(B)连接的边缘为2,则到B到A的距离将为6 + 2 = 8.如果此距离小于先前记录的距离,则覆盖距离
4.当我们考虑当前节点的所有邻居时,将其标记为已访问.将不再检查受访节点; 现在记录的距离是最终的,最小的
我想我是在正确的轨道上,我只是坚持如何说'从节点开始,从源到节点获取长度,如果长度较小,覆盖以前的值,然后移动到下一个节点
我还用字典来存储网络.
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':{}
}
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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')
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