TikZ改变矩阵单元格颜色背景

Question

我有一个可以生成以下内容的 TikZ 代码：

使用类似于以下代码片段的内容。可以从问题的答案中获得与此代码片段相对应的完整代码tikz-双块矩阵：

    \pgfmathparse{\y-\x<1?"draw":""}
    \node[\pgfmathresult](mynode-\x-\y) at (\y,-\x) {$\times$};

但我希望为这些选定的节点设置以下主题。此示例的代码可在另一个问题的答案中找到tikz-标记矩阵中的多个块并使用矩阵数组定义，而不是使用 TikZ 绘制所有内容：

我如何使用 TikZ 做到这一点？这意味着基本上没有边框线，而是将正方形的背景更改为特定的颜色...

更新：以下代码和输出是独立的，并显示了我想要添加背景颜色变化的现有 Tikz 图。

\documentclass{standalone}
    \usepackage{tikz}

    \usetikzlibrary{calc,fit,matrix,arrows,automata,positioning}
\begin{document}
\begin{tikzpicture}
[zeros/.style={},transform shape]
\foreach \y in {1,...,4}{
    \foreach \x in {1,...,5}{
    \ifnum\x>\numexpr\y+1
    \node[zeros] (mynode-\x-\y) at (\y,-\x) {0};
    \else
    \pgfmathparse{\y-\x<1?"draw":""}
    \node[\pgfmathresult](mynode-\x-\y) at (\y,-\x) {$\times$};
    \ifnum\x<5
    \ifnum\x<\y\draw[<-] (mynode-\x-\y) -- +(0,-0.7);\fi
    \ifnum\x=\y\draw[<-] (mynode-\x-\y) -- +(0,-0.7);\fi
    \ifnum\y<4\draw[<-] (mynode-\x-\y) -- +(0.7,-0.7);\fi
    \fi
    \fi
    }
}
\node[fit=(mynode-1-1)(mynode-5-4),draw,thick,red,label=90:{Conquer diagonal}]{};

\begin{scope}[xshift=5.2cm]
    \foreach \x[remember=\x as \lastx] in {1,...,5}{
        \node (myvecs1-\x) at (0,-\x) {$\times$};
        \node (myvecs2-\x) at (1,-\x) {$\times$};
        \node (myvecs3-\x) at (2,-\x) {$\times$};
        \node (myvecs4-\x) at (3,-\x) {$\times$};
        \node (myvecs5-\x) at (4,-\x) {$\times$};
        \node (myvecsn-\x) at (5,-\x) {$\times$};
        \ifnum\x>1
        \draw[->] (myvecs1-\x) -- (myvecs1-\lastx);
        \draw[->] (myvecs2-\x) -- (myvecs2-\lastx);
        \draw[->] (myvecs3-\x) -- (myvecs3-\lastx);
        \draw[->] (myvecs4-\x) -- (myvecs4-\lastx);
        \draw[->] (myvecs5-\x) -- (myvecs5-\lastx);
        \draw[->] (myvecsn-\x) -- (myvecsn-\lastx);
        \fi
        }
\node[fit=(myvecs1-1)(myvecsn-5),draw,thick,green,label=90:{Apply as deep as possible}] {};
\end{scope}
\end{tikzpicture}
\end{document}

输出结果为：

Answer 1

你可以这样做：

代码

\documentclass[tikz]{standalone}
\usetikzlibrary{fit}

\begin{document}

\begin{tikzpicture}
[   zeros/.style={},
    transform shape
]
\foreach \y in {1,...,4}
{   \foreach \x in {1,...,5}{
  \ifnum\x>\numexpr\y+1
    \node[zeros] (mynode-\x-\y) at (\y,-\x) {0};
  \else
    \node(mynode-\x-\y) at (\y,-\x) {$\times$};
    \ifnum\x<5
        \ifnum\x=\y\fill[red,opacity=0.1] (mynode-\x-\y.south west) rectangle (mynode-\x-\y.north east);\fi
    \fi
    \ifnum\y=\numexpr\x-1\fill[opacity=0.1] (mynode-\x-\y.south west) rectangle (mynode-\x-\y.north east);\fi
  \fi
  }
}
\node[fit=(mynode-1-1)(mynode-5-4),draw,thick,red,label=90:{Conquer diagonal}]{};
\end{tikzpicture}

\end{document}

结果