python实现狄克斯特拉算法实例_Python

数据结构

1、路由信息

dictRoute = {}
dictRoute[nodeId] = {}
dictRoute[nodeId][nebrId] = distance
操作：
①根据nodeId找到该node的路由信息
②根据nebrId找到某一条路由的距离

2、节点信息

dictNode = {}
dictNode[nodeId] = [shortDis, fatherId, bIsCheck]
操作：
①找到nodes中最短距离的节点
②查找节点的shortDis，根据情况更新shortDis、fatherId
③检查过的节点，更新bIsCheck

功能实现

/* 找到最短距离节点的Id，已经检查的不计算在内 */
def FindShortNodeId(dictNode):
return shortNodeId

/* dikstra算法流程 */
1、找到最短距离节点Id，并标记已检查过（如果节点Id不存在，表示查找完成）
2、得到最短距离节点的距离
3、轮询最短距离节点的邻居节点
4、计算邻居节点的新距离、得到原最短距离，进行比较
5、如果新距离 < 原距离，则更新邻居节点最短距离
概括为两步：步骤1 （1）- 找到当前最短距离节点
步骤2（2~5） - 更新最短距离节点邻居节点信息

代码实现

import os
import sys

'''
信息输入：
1、节点数目、路由数目
2、路由信息 
3、开始节点、结束节点
'''
nodeNum = 0 # 节点数目
routeNum = 0 # 路由数目
listRoute = [] # 临时存储输入的路由信息
listNodeId = []# 临时存储节点id 

nodeIdStart = ''
nodeIdEnd = ''
dictRoute = {} # 解析后的路由信息
dictNode = {} # 节点信息
# 输入节点数目、路由数目
strInput = input()
list0 = strInput.split(' ')
nodeNum = int(list0[0])
routeNum = int(list0[1])

# 输入路由信息
for index in range(routeNum):
 strInput = input()
 listRoute.append(strInput)
 
# 输入开始节点、结束节点
strInput = input()
list0 = strInput.split(' ')
nodeIdStart = list0[0]
nodeIdEnd = list0[1]

# 解析得到节点Id
listNodeId.append(nodeIdStart)
listNodeId.append(nodeIdEnd)
for index in listRoute:
 list0 = index.split(' ')
 nodeIdA = list0[0]
 nodeIdB = list0[1]
 if nodeIdA not in listNodeId:
  listNodeId.append(nodeIdA) 
 if nodeIdB not in listNodeId:
  listNodeId.append(nodeIdB) 

# 初始化路由信息字典、节点信息字典
for nodeId in listNodeId:
 # 节点字典信息
 dictNode[nodeId] = [10000, '', False] # 最短距离、父节点、是否检查过
 # 每个路由字典创建
 dictRoute[nodeId] = {}
dictNode[nodeIdStart][0] = 0

# 初始化路由信息
for index in listRoute:
 list0 = index.split(' ')
 nodeIdA = list0[0]
 nodeIdB = list0[1]
 dictRoute[nodeIdA][nodeIdB] = int(list0[2])
 dictRoute[nodeIdB][nodeIdA] = int(list0[2])
 
# 打印输入信息
def PrintInputInfo():
 print('nodeNum routeNum:')
 print(str(nodeNum) + ' ' + str(routeNum))
 print('nodeStart nodeEnd')
 print(nodeIdStart+' '+nodeIdEnd)
 print('route info:')
 for nodeId in dictRoute.keys():
  for nebrId in dictRoute[nodeId].keys():
   print(nodeId+'->'+nebrId+' = '+str(dictRoute[nodeId][nebrId]))
 print('node info:')
 for nodeId in dictNode.keys():
  print(nodeId+':'+str(dictNode[nodeId][0])+' '+dictNode[nodeId][1]+' '+str(dictNode[nodeId][2]))

#PrintInputInfo()

'''
狄克斯特拉实现
'''
# 找到最短距离节点id
def FindShortNodeId(dictNode):
 shortNodeId = ''
 shortDis = 10000
 for nodeId in dictNode.keys():
  if dictNode[nodeId][0] < shortDis and dictNode[nodeId][2] == False:
   shortNodeId = nodeId
   shortDis = dictNode[nodeId][0]
 return shortNodeId
 
# 狄克斯特拉算法
shortNodeId = FindShortNodeId(dictNode)
while shortNodeId:
 if shortNodeId == nodeIdEnd:
  break;
 dictNode[shortNodeId][2] = True
 shortDis = dictNode[shortNodeId][0]
 for nebrId in dictRoute[shortNodeId].keys():
  newDis = dictRoute[shortNodeId][nebrId] + shortDis
  if newDis < dictNode[nebrId][0]:
   dictNode[nebrId][0] = newDis
   dictNode[nebrId][1] = shortNodeId
 shortNodeId = FindShortNodeId(dictNode)
 
# 打印结果
listRst = []
nodeId = nodeIdEnd
while nodeId:
 listRst.append(nodeId)
 nodeId = dictNode[nodeId][1]
listRst.reverse()

strRst = ''
for nodeId in listRst:
 if nodeId == listRst[-1]:
  strRst += nodeId
 else:
  strRst += nodeId + '->'

if dictNode[nodeIdEnd][1] == '':
 print('cant reach '+nodeIdEnd)
else:
 print(strRst)
 print(dictNode[nodeIdEnd][0])