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以一种"廉价"的方式获取页面的标题(downmoon)

有朋友问到：“我需要得到一个webpage的title，因为这个title一般都比较靠前，只要取得html的前面少量内容就可以了。因为要取得很多个页面的title,如果下载整个htmlcode，肯定比较浪费时间。.net好像没有现成的类可以干这件事情(取得部分html)，我应该如何去实现？”一种比较"廉价"(即较小成本条件下)的解决思路：第一步：取出含有页面title部分的最小集合。这是“廉价”的关键！第二步：用正则表达式取出和</title</span> <p class="auth-span"> <span class="muted"><i class="fa fa-user"></i><span class="zj-icon-text">系统</span></span> <span class="muted"><i class="fa fa-clock-o"></i><span class="zj-icon-text">2019-08-29 22:21:16</span></span> <span class="muted"><i class="fa fa-eye"></i><span class="zj-icon-text">2647</span></span>  </p> </article><article class="excerpt"> <header> <a title="编程技术" class="label label-important zj-del-target" href="/category/30">编程技术<i class="label-arrow"></i></a> <h2> <a class="zj-titles zj-del-target" href="/article/12812" title="回溯法之一---算法框架及基础" target="_blank" onclick="return recordId('',)">回溯法之一---算法框架及基础 </a> </h2> </header> <div class="focus"> <a class="zj-del-target" href="/article/12812" target="_blank"><img style="width: 200px; height: 123px;" class="thumb" src="/Public/static/images/article/19.jpg" alt="" title=""></a> </div> <span class="note" onclick="if (jq.browser.versions.mobile) {window.location.href='/article/12812';}else{window.open('/article/12812');}" style="cursor: pointer;">回溯法其实也是一种搜索算法，它可以方便的搜索解空间。回溯法解题通常可以从以下三步入手：1、针对问题，定义解空间2、确定易于搜索的解空间结构3、以深度优先的方式搜索解空间，并在搜索的过程中进行剪枝回溯法通常在解空间树上进行搜索，而解空间树通常有子集树和排列树。针对这两个问题，算法的框架基本如下：用回溯法搜索子集合树的一般框架：Cpp代码voidbacktrack(intt){if(t>n)output(x);else{for(inti=f(n,t);i<=g</span> <p class="auth-span"> <span class="muted"><i class="fa fa-user"></i><span class="zj-icon-text">系统</span></span> <span class="muted"><i class="fa fa-clock-o"></i><span class="zj-icon-text">2019-08-12 09:30:35</span></span> <span class="muted"><i class="fa fa-eye"></i><span class="zj-icon-text">2647</span></span>  </p> </article><article class="excerpt"> <header> <a title="数据库相关" class="label label-important zj-del-target" href="/category/32">数据库相关<i class="label-arrow"></i></a> <h2> <a class="zj-titles zj-del-target" href="/article/8293" title="uva 10051 Tower of Cubes(DAG最长路）" target="_blank" onclick="return recordId('',)">uva 10051 Tower of Cubes(DAG最长路） </a> </h2> </header> <div class="focus"> <a class="zj-del-target" href="/article/8293" target="_blank"><img style="width: 200px; height: 123px;" class="thumb" src="/Public/static/images/article/14.jpg" alt="" title=""></a> </div> <span class="note" onclick="if (jq.browser.versions.mobile) {window.location.href='/article/8293';}else{window.open('/article/8293');}" style="cursor: pointer;">题目连接：10051-TowerofCubes题目大意：有n个正方体，从序号1~n，对应的每个立方体的6个面分别有它的颜色（用数字给出），现在想要将立方体堆成塔，并且上面的立方体的序号要小于下面立方体的序号，相邻的面颜色必须相同。输出最高值和路径。解题思路：因为立方体可以旋转，所以一个序号的立方体对应这6种不同的摆放方式，可以将问题理解成DAG最长路问题，只是搜索范围是从i+1开始到n。然后记录路径要开两个2维数组。路径不唯一，随便输出一条。#includ</span> <p class="auth-span"> <span class="muted"><i class="fa fa-user"></i><span class="zj-icon-text">系统</span></span> <span class="muted"><i class="fa fa-clock-o"></i><span class="zj-icon-text">2019-08-12 01:55:00</span></span> <span class="muted"><i class="fa fa-eye"></i><span class="zj-icon-text">2647</span></span>  </p> </article><article class="excerpt"> <header> <a title="Oracle" class="label label-important zj-del-target" href="/category/45">Oracle<i class="label-arrow"></i></a> <h2> <a class="zj-titles zj-del-target" href="/article/6016" title="学习Oracle日记（二）" target="_blank" onclick="return recordId('',)">学习Oracle日记（二） </a> </h2> </header> <div class="focus"> <a class="zj-del-target" href="/article/6016" target="_blank"><img style="width: 200px; height: 123px;" class="thumb" src="/Public/static/images/article/4.jpg" alt="" title=""></a> </div> <span class="note" onclick="if (jq.browser.versions.mobile) {window.location.href='/article/6016';}else{window.open('/article/6016');}" style="cursor: pointer;">＃１.查询语句中使用函数@1.round(d1,d2)用于数字的四舍五入d1表示数字d2表示小数点后的位数（d2不写默认为零）@2.trunc(d1,d2)截取去尾法@3.日期函数sysdateselectsysdatefromdual;----获取系统当前时间@4.日期数据相减，得到两个日期之间的天数差不足一天用小数表示@5.months_between(d1,d2)得到d1,d2两个日期数据之间所差的月份小数表示@6.计算d2个月之前的时间点add_m</span> <p class="auth-span"> <span class="muted"><i class="fa fa-user"></i><span class="zj-icon-text">系统</span></span> <span class="muted"><i class="fa fa-clock-o"></i><span class="zj-icon-text">2019-08-12 01:52:38</span></span> <span class="muted"><i class="fa fa-eye"></i><span class="zj-icon-text">2647</span></span>  </p> </article><article class="excerpt"> <header> <a title="数据库相关" class="label label-important zj-del-target" href="/category/32">数据库相关<i class="label-arrow"></i></a> <h2> <a class="zj-titles zj-del-target" href="/article/5496" title="委托事件简写与原来写法 (临时函数)转" target="_blank" onclick="return recordId('',)">委托事件简写与原来写法 (临时函数)转 </a> </h2> </header> <div class="focus"> <a class="zj-del-target" href="/article/5496" target="_blank"><img style="width: 200px; height: 123px;" class="thumb" src="/Public/static/images/article/13.jpg" alt="" title=""></a> </div> <span class="note" onclick="if (jq.browser.versions.mobile) {window.location.href='/article/5496';}else{window.open('/article/5496');}" style="cursor: pointer;">publicMainPage(){this.InitializeComponent();//简写this.Loaded+=(sender,e)=>{MessageBox.Show("d");MessageBox.Show("ss");};//原来写法this.Loaded+=newRoutedEventHandler(MainPage_Loaded);}voidMainPage_Loaded(objectsender,RoutedEventArgse){/</span> <p class="auth-span"> <span class="muted"><i class="fa fa-user"></i><span class="zj-icon-text">系统</span></span> <span class="muted"><i class="fa fa-clock-o"></i><span class="zj-icon-text">2019-08-12 01:52:04</span></span> <span class="muted"><i class="fa fa-eye"></i><span class="zj-icon-text">2647</span></span>  </p> </article><div class="pagination" style="display:;"> <ul><li><a href="/search/email/1/p/1">首页</a></li> <li class="prev-page"><a 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