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自我介绍1

Good morning. Dear Teachers I am glad to be here for this interview.

  • good 的 d不要发音,不是 猫宁 or glad 嘴张大 ‘哎’

My name is Zhanzeye, 21 years old,I come from AnHui,Anqing. I am going to graduate from the information and computer engineering department of Northeast Forestry University in july, 2020.

In the past three years I have been preparing for the postgraduate recommendation.

Now all my hard work has got a result since I have a chance to be here.

  • got 读 \^\ 和a 连读 发d

So I just feel a little bit nervous, because this is my first time to have such an interview.

  • just -> t ; bit -> t (字母t 不发音) because -> ‘cos (课 zi) an interview 降调

In the past three years, I have learnt some basic knowledge of software engineering.

For example, from the course of software architecture , I got to know software design pattern.

  • for example r 和 e 连读

When I learnt more about this field. I became more interested in it.

  • in的词尾[n]和it的词首[ɪ]连读为[nɪ]

Apart from textbooks, I borrowed many relevant ones from our library to read, such as data structure and network of computer

I really enjoy my journey when reading and absorbing the information from these fascinating works.

  • journey 前停顿 升调 the 在 information 读 zi

In addition to reading books, I also like browsing through some journals in 专业领域 on a daily basis so that I can keep an eye on the latest news and findings in my area.

  • keep an eye 连读

My passion in software is the reason why I would like to pursue my further study in this field. Owing to my

  • pursue => I miss you => ensure \s\ 和 \j\

diligence and persistence, I could concentrate on my study and I believe that Icould eventually succeed.

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概述

Floyd 算法是一种利用动态规划的思想、计算给定的带权图中任意两个顶点之间最短路径的算法。相比于重复执行多次单源最短路算法,Floyd 具有高效、代码简短的优势,在解决图论最短路题目时比较常用。

我们dp[k][i][j] 表示 i 到 j 能经过 1~ k 的点的最短路。那么实际上dp[0][i][j] 就是原图,如果 i, j之间存在边,那么 i , j之间不经过任何点的最短路就是边长,否则,i, j之间的最短路为无穷大。

那么对于 i, j 之间经过 1 ~ k 的最短路 dp[k][i][j] 可以通过经过 1 ~ k - 1 的最短路转移过来。

  • 如果不经过第 k个点,那么就是 dp[k - 1][i][j]
  • 如果经过第 k 个点,那么就是 dp[k - 1][i][k] + dp[k - 1][k][j]

所以就有转移

$\displaystyle dp[k][i][j] = min(dp[k - 1][i][j], dp[k - 1][i][k] + dp[k - 1][k][j])$

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2019 - 06 - 29

Policy in the spotlight as IP theft, net addiction and supercomputing feature

  1. giant

    n. 巨人;伟人;大公司

    adj. 巨大的;巨人般的;伟大的

  1. balk at

    回避,畏缩

  1. bounce

    v. 弹起;反弹;蹦跳

    n. 弹跳;弹力;活力;(突然的)上升

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第一课
  1. 替换 indicate, suggest, fear

    be indicative of
    be suggestive of
    be fearful of

    The clouds are indicative of the coming of rain.

    I **am suggestive of ** trying once more.
    We are fearful of time!

  1. 替换 cause
    give rise to
    lead to
    result in
    trigger
    A bad tooth can trigger/give rise to/lead to acute pain.

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第一课
  • impact n. 影响
    effect, influence
  • issue n. 发表[声明], 发出(命令), 争论的问题
    officially make a statement; give an order; officially produce;
    subject
  • consequence n. 后果
    result
  • transition v. 过度;转变
    change to
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SPFA 算法

SPFA(Shortest Path Faster Algorithm)算法是单源最短路径的一种算法,通常被认为是 Bellman-ford 算法的队列优化,在代码形式上接近于宽度优先搜索 BFS,是一个在实践中非常高效的单源最短路算法。

在 SPFA 算法中,使用 d[i] 表示从源点到顶点 i 的最短路,额外用一个队列来保存即将进行拓展的顶点列表,并用 inq[i] 来标识顶点 i 是不是在队列中。

  1. 初始队列中仅包含源点,且源点 s 的 d[s] = 0 ( 且其他的为INF )。
  2. 取出队列头顶点u,扫描从顶点u出发的每条边,设每条边的另一端为v,边 <u,v>权值为 w,若 d[u] + w < d[v],则
    • 将 d[v] 修改为 d[u] + w
    • 若v不在队列中,则
      • 将 v 入队
  3. 重复步骤 2 直到队列为空

最终 d 数组就是从源点出发到每个顶点的最短路距离。如果一个顶点从没有入队,则说明没有从源点到该顶点的路径。

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第一课

The past several years have witnessed a booming economy in China. However, it is increasingly difficult for college undergraduates to find a decent job.

  • booming adj. 生机勃勃
  • decent adj. 得体的, 相当好的

There are several factors contributing to this social phenomenon. To begin with, the job market is getting gloomy and competition is becoming fierce.

  • gloomy adj. 忧郁的,低迷的

Moreover, the expectations of employment play an important part. A large number of graduates tend to give first priority to earnings and aim for well-paid position in foreign companies or big banks.

  • priority n. 优先

Last but not least, many employers do require job applicants to have a degree relate to the job.

  • degree n.学历

One has to have additional education to get an edge. Education is indeed a worthwhile investment in the future.

  • edge n. 优势
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