Python小程序：用广度优先搜索算法查询两个url之间的最短路径

源自人工智能一次作业中的一道题

本人Python小白啊，真的要好好努力了。话不多说，直接上代码。

#encoding:UTF-8
import time
from lxml import etree
import urllib2
import Queue
import requests
requests.packages.urllib3.disable_warnings()


def findUrl(url):  # 查找Url对应页面包含的子url
    try:
        links = []
        res = requests.get(url, timeout=0.6)  # 获取requests发出get请求后的response对象
        html = etree.HTML(res.text)           # etree解析响应体的字符串形式
        newurls = html.findall('.//a')        # 找到所有的a标签
        for newURL in newurls:
            href = newURL.get('href')         # 获取href链接
            links.append(href)
        return links
    except Exception as e:
        pass

def printPath(parents, startUrl, targetUrl):  # 寻找从初始url到目标url的路径
    try:
        print "\nThe path from %s to %s: " % (startUrl, targetUrl)
        path = [targetUrl]
        parent = parents[targetUrl]           # 寻找父亲
        while bool(parent) == True:
            path.append(parent)
            parent = parents[parent]
        path = path[::-1]                     # 列表反转
        print "\n-> ".join(path)                
    except Exception as e:
        pass

def search(startUrl, targetUrl):
    queue = Queue.Queue()                # 队列 存储 未访问的url
    visited = set()                      # 集合 存储 访问过的url
    parents = dict()                     # 字典 存储 父url
    parents[startUrl] = None             # 起始url为祖先 
    queue.put(startUrl)                  # BFS开始前先将源url推进队列
    visited.add(startUrl)                # 初始节点标记访问

    while (queue.empty() == False):      # 队列非空
        try:
            curentUrl = queue.get()      # 取出队列首部并pop掉
            print('search in %s ...' % curentUrl)

            # 找出当前url的所有子url
            urlLink = findUrl(curentUrl)
            if urlLink:
                for url in urlLink:       
                    parents[url] = curentUrl     # 记录当前url与子url的对应关系
                    if (url == targetUrl):
                        print('find %s successfully\n' % targetUrl)
                        printPath(parents, startUrl, targetUrl)   # 打印路径
                        return
                    if (url not in visited):     # 如果子url还未访问过，推进队列并标记访问
                        queue.put(url)
                        visited.add(url)
        except Exception as e:
            pass



if __name__ == '__main__':

  startTime = time.time()

  search('http://helpdesk.sysu.edu.cn/', 'http://tv.sysu.edu.cn/')

  print ("\nCost time: %f s" % (time.time() - startTime))

查找过程如下：

查询结果如下：

提示成功找到目标url, 打印最短路径，并显示查询时间。