我正在绘制一个具有不同状态名称的大型马尔可夫链；一些状态包含数字，例如 (0,2)，而一些状态则命名为 (S-1,R-1)。我正在尝试找到一种方法，让所有状态的字体大小都一致。具有 S 和 R 的状态比其他状态大。有没有办法让状态大小和字体大小一致？我的代码没有转换，如下所示
\documentclass{document}
\usepackage{tikz}
\usetikzlibrary{positioning}
\usetikzlibrary{arrows}
\usetikzlibrary{chains,shapes.multipart}
\usetikzlibrary{shapes}
\usetikzlibrary{automata,}
\setlength{\jot}{7pt}
\usepackage{multirow}
\usepackage{latexsym}
\usepackage[utf8]{inputenc}
%\usepackage[dvipsnames]{xcolor}
\definecolor {processblue}{cmyk}{0,0,0,0.17}
\definecolor{light-gray}{gray}{0.85}
\newcommand*{\grayemph}[1]{%
\tikz[baseline=(X.base)] \node[rectangle, fill=light-gray, rounded corners, inner sep=1.8mm] (X) {#1};%
}
\begin{document}
\begin{figure*}[t]
\centering
\begin{tikzpicture}
\tikzset{
>=stealth',
node distance=1.5cm and 1.7cm ,on grid,
every text node part/.style={align=center}
state/.style={minimum size=30pt,font=\small,circle,draw},
dots/.style={state,draw=none},
edge/.style={->},
trans/.style={font=\footnotesize,above=2mm},
reflexive/.style={out=120,in=60,looseness=5,relative},
}
\node[state] (C){$0,0$};
\node[state] (v1) [right =of C] {$1,0$};
\node[state] (v2) [right =of v1] {$2,0$};
\node[state] (v3) [right =of v2] {$\cdots$};
\node[state] (v4) [right =of v3] {\begin{scriptsize}S-1, {0}\end{scriptsize}};
\node[state] (v5) [right =of v4] {$S,0$};
\node[state] (v6) [above =of C] {$1,0$};
\node[state] (v7) [right =of v6] {$1,1$};
\node[state] (v8) [right =of v7] {$2,1$};
\node[state] (v9) [right =of v8] {$\cdots$};
\node[state] (v10) [right =of v9] {\begin{scriptsize}S-1, {0}\end{scriptsize}};;
\node[state] (v11) [right =of v10] {$S,1$};
\node[state] (v12) [right =of v11] {\begin{scriptsize}S+1, {0}\end{scriptsize}};;
\node[state] (v13) [above =of v6] {$\vdots$};
\node[state] (v14) [above =of v7] {$\vdots$};
\node[state] (v15) [above =of v8] {$\vdots$};
\node[state] (v16) [above =of v9] {$\vdots$};
\node[state] (v17) [above =of v10] {$\vdots$};
\node[state] (v18) [above =of v11] {$\vdots$};
\node[state] (v19) [right =of v18] {$\ddots$};
\node[state] (v20) [above =of v13] {\begin{scriptsize}0,R-1\end{scriptsize}};
\node[state] (v21) [right =of v20] {\begin{scriptsize}1,R-1\end{scriptsize}};
\node[state] (v22) [right =of v21] {\begin{scriptsize}2,R-1\end{scriptsize}};
\node[state] (v23) [right =of v22] {$\cdots$};
\node[state] (v24) [right =of v23] {\begin{scriptsize}S-1, {R-1}\end{scriptsize}};;
\node[state] (v25) [right =of v24] {\begin{scriptsize}S, {R-1}\end{scriptsize}};
\node[state] (v26) [right =of v25] {\begin{scriptsize}S+1, {R-1}\end{scriptsize}};
\node[state] (v27) [right =of v26] {$\cdots$};
\node[state] (v28) [right =of v27] {\begin{scriptsize}S+R-1,R-1\end{scriptsize}};
\end{tikzpicture}
\caption{The Markov chain mode;.}
\end{figure*}
\end{document}
答案1
有些文本在里面scriptsize，而其他文本（用 $...$ 括起来的文本）不在里面。这导致字体大小不同。节点minimum size规范必须大于文本最长的节点，以便所有节点的大小相同（在此代码中，右上方节点的文本最长）。所以我也更改了最小尺寸规范。
再次，正如我之前在问题评论中所说，缺少的逗号似乎掩盖了规范state/.style。所以我在那里添加了一个逗号。只有那时 Tikz 才会读取该值。
\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{positioning}
\usetikzlibrary{arrows}
\usetikzlibrary{chains,shapes.multipart}
\usetikzlibrary{shapes}
\usetikzlibrary{automata}
\begin{document}
\begin{tikzpicture}
\tikzset{
>=stealth',
node distance=2.5cm and 2.5cm ,on grid,
every text node part/.style={align=center},
state/.style={minimum width=2cm,
font=\small,
draw,
circle},
}
\node[state] (C){$0,0$};
\node[state] (v1) [right =of C] {$1,0$};
\node[state] (v2) [right =of v1] {$2,0$};
\node[state] (v3) [right =of v2] {$\cdots$};
\node[state] (v4) [right =of v3] {\begin{scriptsize}S-1, {0}\end{scriptsize}};
\node[state] (v5) [right =of v4]
{\begin{scriptsize}S,0\end{scriptsize}};
\node[state] (v6) [above =of C] {$1,0$};
\node[state] (v7) [right =of v6] {$1,1$};
\node[state] (v8) [right =of v7] {$2,1$};
\node[state] (v9) [right =of v8] {$\cdots$};
\node[state] (v10) [right =of v9] {\begin{scriptsize}S-1, {0}\end{scriptsize}};;
\node[state] (v11) [right =of v10] {\begin{scriptsize}S,1\end{scriptsize}};
\node[state] (v12) [right =of v11] {\begin{scriptsize}S+1, {0}\end{scriptsize}};;
\node[state] (v13) [above =of v6] {$\vdots$};
\node[state] (v14) [above =of v7] {$\vdots$};
\node[state] (v15) [above =of v8] {$\vdots$};
\node[state] (v16) [above =of v9] {$\vdots$};
\node[state] (v17) [above =of v10] {$\vdots$};
\node[state] (v18) [above =of v11] {$\vdots$};
\node[state] (v19) [right =of v18] {$\ddots$};
\node[state] (v20) [above =of v13] {\begin{scriptsize}0,R-1\end{scriptsize}};
\node[state] (v21) [right =of v20] {\begin{scriptsize}1,R-1\end{scriptsize}};
\node[state] (v22) [right =of v21] {\begin{scriptsize}2,R-1\end{scriptsize}};
\node[state] (v23) [right =of v22] {$\cdots$};
\node[state] (v24) [right =of v23] {\begin{scriptsize}S-1, {R-1}\end{scriptsize}};;
\node[state] (v25) [right =of v24] {\begin{scriptsize}S, {R-1}\end{scriptsize}};
\node[state] (v26) [right =of v25] {\begin{scriptsize}S+1, {R-1}\end{scriptsize}};
\node[state] (v27) [right =of v26] {$\cdots$};
\node[state] (v28) [right =of v27] {\begin{scriptsize}S+R-1,R-1\end{scriptsize}};
\end{tikzpicture}
\end{document}