TikZ - 子样式 + 相对坐标

Question

我有两个问题需要用以下 TiKz 代码解决。

1. 我想使用相对坐标，例如获得更宽的蓝色矩形。我尝试过但++没有成功。
2. 我的所有样式都有通用定义。有没有办法使用一种可供我所有样式使用的子样式？

我该如何解决这个问题？

\documentclass[12pt]{amsart}
\usepackage{tikz}
\usetikzlibrary{arrows, matrix, positioning}

\tikzset{
    % Good spacing hack (the ghost mode)
    gs/.style={
        rectangle,
        thick,
        text width=3em,
        align=center,
        draw=white,
        rounded corners,
        minimum height=2em
    },    % Focus effect
    fe/.style={
        rectangle,
        thick,
        rounded corners,
        minimum height=2em,
        text width=8em,
        align=center,
        draw=red,
        text=magenta
    },
    % Focus effect Bis
    feb/.style={
        rectangle,
        thick,
        rounded corners,
        minimum height=2em,
        text width=8em,
        align=center,
        draw=blue,
        text=violet
    },
    % Long focus effect
    lfe/.style={
        rectangle,
        thick,
        rounded corners,
        minimum height=2em,
        text width=12em,
        align=center,
        draw=red,
        text=magenta
    },
    % Long focus effect bis
    lfeb/.style={
        rectangle,
        thick,
        rounded corners,
        minimum height=2em,
        text width=12em,
        align=center,
        draw=blue,
        text=violet
    },
}


\begin{document}

\begin{tikzpicture}
    \matrix[matrix of nodes,
        row sep    = 1em,
        column sep = 1.5em,
    ](mat){
          & |[gs]|$a$ & |[gs]|$X$ \\
|[gs]|$q$ & |[gs]|$b$ & |[gs]|$Y$  \\
          & |[gs]|$r$ & |[gs]|$Z$  \\
    };

    \draw[fe] (mat-1-2.north west) -- (mat-1-3.north east)
           -- (mat-2-3.south east) -- (mat-2-2.south west)
           -- (mat-1-2.north west);

    \draw[red,-triangle 60] (mat-1-3.north east) -- ++ (0,-2.5em) |- ++ (3em,0)
               node[pos=2.05, lfe] (f){$aY - bX = d$};

    \draw[feb] (mat-2-2.north west) -- (mat-2-3.north east)
           -- (mat-3-3.south east) -- (mat-3-2.south west)
           -- (mat-2-2.north west);

    \draw[blue,-triangle 60] (mat-2-3.north east) -- ++ (0,-2.5em) |- ++ (3em,0)
               node[pos=2.05, lfeb] (f){$bZ - rY = e$};
\end{tikzpicture}

\end{document}

Answer 1

您可以直接用 绘制矩形fit。要使它们变大，您可以增加inner sep。

\documentclass[12pt]{amsart}
\usepackage{tikz}
\usetikzlibrary{arrows, matrix, positioning,fit}

\tikzset{
    % Good spacing hack (the ghost mode)
     gs/.style={
        rectangle,
        thick,
        text width=3em,
        align=center,
        draw=white,
        rounded corners,
        minimum height=2em
    },    %
    rc/.style={rectangle,
        thick,rounded corners,draw},
    lfe/.style={rc,
        text width=12em,
        align=center,
        text=magenta
    },
    % Long focus effect bis
    lfeb/.style={rc,
        text width=12em,
        align=center,
        text=violet
    },
}


\begin{document}

\begin{tikzpicture}
    \matrix[matrix of nodes,nodes=gs,
        row sep    = 1em,
        column sep = 1.5em,
    ](mat){
          & $a$ & $X$ \\
$q$ & $b$ & $Y$  \\
          & $r$ & $Z$  \\
    };

    \node[fit=(mat-1-2) (mat-2-3),inner xsep=1em,rc,red] (F1){};

    \draw[red,-triangle 60] (F1.east) -- ++ (2.5em,0)
               node[right, lfe] (f){$aY - bX = d$};

    \node[inner xsep=2em,fit=(mat-2-2) (mat-3-3),rc,blue] (F2){};

    \draw[blue,-triangle 60] (F2.east) -- ++ (2.5em,0)
               node[right,lfeb] (f){$bZ - rY = e$};
\end{tikzpicture}

\end{document}