答案1
使用该tikz-cd
包的一种可能性:
\documentclass{memoir}
\usepackage{tikz-cd}
\begin{document}
\begin{center}
\begin{tabular}{c @{\hspace{4em}} l}
\begin{tikzcd}[row sep=small, baseline=(current bounding box.center)]
& \rlap{aaa} \\
\llap{x} \ar[ru]\ar[rd] & \\
& \rlap{bbb}
\end{tikzcd} & text \\
\begin{tikzcd}[row sep=small, baseline=(current bounding box.center)]
& \rlap{aaa} \\
\llap{y} \ar[ru]\ar[rd] & \\
& \rlap{bbb}
\end{tikzcd} & text \\
\begin{tikzcd}
\llap{longtext} & \rlap{ccc}
\end{tikzcd} & text
\end{tabular}
\end{center}
\end{document}
答案2
一种简单灵活的方法是使用普通的 TikZ。
\documentclass{memoir}
\usepackage{tikz}
\usepackage{amsmath,amssymb}
\begin{document}
\begin{center}
\begin{tikzpicture}[>=stealth]
\path[left]
(0,0) node (A) {\r{a}ben}
++(0,-1.6) node (L) {lukket}
++(0,-1.2) node (D) {dobbeltlukket};
\path[right] (A.east)
+(1,.5) node (A1) {ka}
+(1,-.5) node (A2) {k\textsuperscript{e}}
+(2.5,0) node{\itshape ingen kort};
\path[right] (L.east)
+(1,.5) node (L1) {kal}
+(1,-.5) node (L2) {k\={a}l}
+(2.5,0) node{\itshape ingen reduceret};
\path (D.east)+(1,0) node[right] {kalt};
\draw[->] (A.east)--(A1);
\draw[->] (A.east)--(A2);
\draw[->] (L.east)--(L1);
\draw[->] (L.east)--(L2);
\end{tikzpicture}
\end{center}
\end{document}