在表格环境中绘制分隔的虚线/实心矩形（子表）

Question

我需要绘制一个如下所示的表格。

与主表格环境中的矩形子表中的符号分组和阴影相反，厚度在这里并不重要。

是否可以使用例如 \tcolorbox{} 来填充虚线框？如何操作？

Answer 1

使用TiKZ matrix并不难做到这一点：

\documentclass[tikz,border=2mm]{standalone} 
\usetikzlibrary{matrix,backgrounds,patterns}

\begin{document}
\begin{tikzpicture}
\matrix (A) [matrix of nodes, column sep=1mm, row sep=1mm] {
0 & 1 & 1 & 0 & 1 & 1 & 0 & 1 \\
1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 \\
1 & 1 & 1 & 1 & 0 & 1 & 1 & 0 \\
0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 & 1 & 1 & 1 & 1 \\};

\draw[ultra thick] (A.south west)--([yshift=2mm]A.north west) coordinate (Ann);
\draw[ultra thick] ([xshift=-2mm]A.north west) coordinate (Aww)--(A.north east);
\path (A-2-3) -- coordinate (aux) (A-3-4);
\draw (aux|-A.south)--(aux|-Ann);
\draw (aux)--(aux-|Aww);
\begin{scope}[on background layer]
\draw[pattern=north west lines] (A-1-2.north west) rectangle (A-1-2.south east);
\draw[pattern=north west lines] (A-1-5.north west) rectangle (A-1-6.south east);
\draw[pattern=north east lines] (A-2-7.north west) rectangle (A-2-8.south east);
\draw[pattern=north west lines] (A-4-2.north west) rectangle (A-5-2.south east);
\draw[pattern=north west lines] (A-4-5.north west) rectangle (A-5-6.south east);
\draw[pattern=north east lines] (A-5-7.north west) rectangle (A-5-8.south east);

\end{scope}

\end{tikzpicture}
\end{document}

Answer 2

这是一个使用 pstricks 的简单解决方案：

\documentclass[x11names]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage{boldline}
\usepackage{pst-node, multido}
\usepackage{auto-pst-pdf}
\newcommand\tleft[1]{\pnode[-1.2ex, 2.5ex]{t#1}}
\newcommand\bright[1]{\pnode[1.2ex, -1.2ex]{b#1}}

\begin{document}

\[ %
\begin{postscript}
\psset{fillstyle=vlines,  hatchwidth=0.3pt,  hatchcolor=IndianRed3}
\renewcommand\arraystretch{1.5}\renewcommand\arraycolsep{8pt}
\begin{array}{cV{8}*{3}{c}|*{5}{c}}
&  &  & \\[-2ex]
\hlineB{8}
& 0 & \tleft{1}1 \bright{1} & 1 & 0 & \tleft{3} 1 & 1 \bright{3} & 0 & 1 \\%
& 1 & 0 & 0 & 1 & 0 & 0 & \tleft{5} 1 & 1 \bright{5} \\%
\cline{1-4}
& 1 & 1 & 1 & 1 & 0 & 1 & 1 & 0  \\%
& 0 &  \tleft{2} 1 & 0 & 0 & \tleft{4} 1 & 1  &  0 & 0 \\%
& 1 & 1 \bright{2} & 0 & 0 &  1 & 1 \bright{4} & \tleft{6}  1 & 1 \bright{6}
\end{array}%
\psset{linecolor=IndianRed3, linejoin=1}
\multido{\i=1+1}{6}{ \psframe(t\i)(b\i)}
\end{postscript}%
 \]

\end{document} 

Answer 3

这是一个解决{NiceArray}方案nicematrix。我用蓝色画了一些阴影，以显示它们被绘制在下面矩阵的系数（如预期的那样）。

\documentclass{article}
\usepackage{nicematrix,tikz}
\usetikzlibrary{fit,patterns}

\begin{document}

$\begin{NiceArray} [rules/width=2pt] {c|cccccccc}
\CodeBefore [create-cell-nodes]
  \begin{tikzpicture} [inner sep = 1pt]
    \begin{scope} [every node/.style = {draw,pattern = north west lines}]
      \node [fit = (2-3)] {} ; 
      \node [fit = (5-3) (6-3)] {} ;
      \node [fit = (2-6) (2-7)] {} ;
      \node [fit = (5-6) (6-7)] {} ;
    \end{scope}
    \begin{scope} [every node/.style = {draw,pattern color=blue, pattern = north east lines}]
      \node [fit = (3-8) (3-9)] {} ; 
      \node [fit = (6-8) (6-9)] {} ; 
    \end{scope}
    \draw (1-|5) -- (last-|5) (4-|1) -- (4-|5) ;
    \end{tikzpicture}
\Body
\\
\hline 
\hspace*{2mm}
& 0 & 1 & 1 & 0 & 1 & 1 & 0 & 1 \\
& 1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 \\
& 1 & 1 & 1 & 1 & 0 & 1 & 1 & 0 \\
& 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\
& 1 & 1 & 0 & 0 & 1 & 1 & 1 & 1 
\end{NiceArray}$

\end{document}

您需要多次编译（因为 PGF/Tikz 节点）。