如何绘制带有箭头的矩形？

Question

我想画出图片中给出的图像。到目前为止，我只能画

\documentclass{beamer}
\usepackage{mathtools}
\usepackage{array}
\usepackage{amsmath}
%\usetheme{Boadilla}
\usetheme{Madrid}
\usepackage{bibentry}
\usepackage[linesnumbered,ruled,vlined]{algorithm2e}
\setbeamertemplate{theorems}[numbered]
\usepackage{tikz}

\usepackage{graphicx}
\usepackage[section]{placeins}
\usepackage[utf8]{inputenc}

 \begin{document}
 \begin{frame}
 \begin{tikzpicture}
 \draw (0,0) -- (3,0) -- (3,4) -- (0,4) -- (0,0);
 \end{tikzpicture}
 \end{frame}

 \end{document}

问题：如何绘制下面的图像？

Answer 1

希望成为 Ti 的起点钾z。

\documentclass{beamer}
\usepackage{mathtools}
\usepackage{array}
\usepackage{amsmath}
%\usetheme{Boadilla}
 \usetheme{Madrid}
 \usepackage{bibentry}
 \usepackage[linesnumbered,ruled,vlined]{algorithm2e}
  \setbeamertemplate{theorems}[numbered]
  \usepackage{tikz}

     \usepackage{graphicx}

    \usepackage[section]{placeins}
   \usepackage[utf8]{inputenc}

 \begin{document}
 \begin{frame}
 \begin{center}
 \begin{tikzpicture}[dot/.style={
draw,
fill,
circle,
minimum size=4pt,
inner sep=0pt, outer sep=0pt,
},
>=stealth]
\node [draw,cyan,ultra thick,rectangle,minimum height=4cm,minimum width=3cm](rect){};
\draw[<->] ([xshift=3mm] rect.north east) -- ([xshift=3mm] rect.south east)node[fill=white,midway]{K};
\draw[ultra thick,<-] ([xshift=3mm] rect.north)--++(90:1)node[right]{text};
\begin{scope}[xshift={-4mm}]
\foreach \y/\k in {1.5/a,1/b,0.5/c}
\node [dot] at (0,\y) (\k) {};
\node at(0,0){\vdots}; 
\foreach \y/\k in {1.5/d,1/e,0.5/f}
\node [dot] at (0,-\y) (\k) {};
\end{scope}
\node at ([xshift=-10mm] rect.west)(m){m};
\draw [->](m) to[bend left](a);
\draw [->](m) to[bend left](b);
\draw [->](m) to[bend right](d);

 \end{tikzpicture}
 \end{center}
 \end{frame}

  \end{document}