设置分区和 tikz

Question

请问，我该如何绘制覆盖节点的彩色矩形，如下图所示，

Answer 1

只是为了好玩，使用条件、绘图定义和嵌套指令优化以前帖子中的代码。

结果：

梅威瑟：

% By J. Leon, Beerware licence is acceptable...
\documentclass[tikz, border=15pt]{standalone}
\usepackage{xcolor} % For custom colors using color pickers.
\definecolor{CustA}{RGB}{228, 127, 226}
\definecolor{CustB}{RGB}{127, 211, 125}
\definecolor{CustC}{RGB}{204, 174, 161}
\definecolor{CustD}{RGB}{248, 185, 126}
\definecolor{CustE}{RGB}{255, 127, 125}
\definecolor{CustF}{RGB}{226, 127, 141}
\usepackage{tikz}
\usetikzlibrary{fit,backgrounds}  

\begin{document}
    \begin{tikzpicture}[
        %Environment Configuration
        font=\Large, %unworn but ready for action
        %Styles
        Line/.style = {% Style for lines
            line width=4mm,
            line cap=round,
            %opacity=0.5
        },
        Area/.style = {% Style for shaded areas
            line width=4mm,
            fill,
            line cap=round,
            rounded corners= 0.5mm,
            %opacity=0.5
        }
        ]
        % Start drawing "the thing..."
        %Create a drawing definition \Mshape(position)[rotatión]{state}
        \def\Mshape(#1)[#2]#3{
            \begin{scope}[shift={(#1)},rotate=#2]
                \foreach \k in {1,...,5}{
                    \draw[fill](72*\k-36:1cm) circle (1mm) coordinate (N\k);
                }
                \begin{scope}[on background layer] % Conditional drawings in the background.
                \ifnum#3=1 \draw[Line,CustA](N3) -- (N4); \fi
                \ifnum#3=2 \draw[Line,CustA](N2) -- (N5); \fi
                \ifnum#3=3 \draw[Area,CustB](N1) -- (N3) -- (N4) -- cycle; \fi
                \ifnum#3=4 \draw[Area,CustB](N5) -- (N1) -- (N2) -- cycle; \fi
                \ifnum#3=5 \draw[Line,CustC](N2) -- (N3); \draw[Line,CustD](N4) -- (N5); \fi
                \ifnum#3=6 \begin{scope}[blend group=multiply]\draw[Line,CustC](N2) -- (N4); \draw[Line,CustD](N3) -- (N5); \end{scope}\fi
                \ifnum#3=7 \draw[Line,CustC](N3) -- (N4); \draw[Line,CustD](N5) -- (N2); \fi
                \ifnum#3=8 \draw[Area,CustE](N2) -- (N3) -- (N4) -- (N5) -- cycle; \fi
                \ifnum#3=9 \draw[Line,CustA](N3) -- (N4); \draw[Area,CustB](N5) -- (N1) -- (N2) -- cycle; \fi
                \ifnum#3=10 \begin{scope}[blend group=multiply]\draw[Line,CustA](N2) -- (N5); \draw[Area,CustB](N1) -- (N3) -- (N4) -- cycle; \end{scope}\fi
                \ifnum#3=11 \draw[Area,CustE](N1) -- (N2) -- (N3) -- (N4) -- (N5) -- cycle; \fi
                \end{scope}
            \end{scope}
        }
        
        \Mshape(0,0)[0]{0} % first drawing at origin
        \foreach \Rot [count=\Yshift from 0] in {1,...,5}{ %copies are made downward and rotating 
            \foreach \case [count=\Xshift from 0]in {1,...,10}{ %copies of the various cases are made forward
                \Mshape(3cm+3cm*\Xshift,-6cm+3cm*\Yshift)[0-72*\Rot]{\case}
            }
        }
        \Mshape(33cm,0)[0]{11} % last drawing with case 11

    \end{tikzpicture}
\end{document}

Answer 2

我对我提供的解决方案并不完全满意，但这是一个开始。
它还没有完成，但你明白了。
不幸的是，它还不够通用，不符合我的口味，我可能会努力改进它。

\documentclass{article}
\usepackage{tikz}

\definecolor{2purple}{RGB}{204,102,255}
\definecolor{3green}{RGB}{0,204,0}

\begin{document}

\begin{tikzpicture}[scale=0.5]
    \tikzset{twopurple/.style={2purple,line width=8pt,rounded corners=2pt,cap=round}}
    \tikzset{threegreen/.style={3green,line width=8pt,rounded corners=2pt,cap=round,fill}}

    \newcommand{\pent}
        {
        \foreach \i in {1,...,5}
                {
                \coordinate (p\i) at (72*\i:\r);
                \filldraw(p\i) circle (\pointradius pt);
                }
        }
            
    \def\r{1}           % pentagon radius
    \def\pointradius{3} % points radius
    
    % First row
    
    \foreach \j/\k [count=\shft from 0] in  
       {1/2,
        2/3,
        3/4,
        4/5,
        5/1}
        {
        \begin{scope}[yshift=-100*\shft]
            \pent           
            \foreach \i in {1,...,5}
                \draw[twopurple] (p\j) -- (p\k);
            \pent           
        \end{scope}
        }

    % Second row
        
    \foreach \j/\k [count=\shft from 0] in  
       {3/5,
        1/4,
        2/5,
        1/3,
        2/4}
        {
        \begin{scope}[xshift=100,yshift=-100*\shft]         
            \pent
            \foreach \i in {1,...,5}
                \draw[twopurple] (p\j) -- (p\k);
            \pent
        \end{scope}
        }
        
    % Third row
        
    \foreach \j/\k/\l [count=\shft from 0] in   
       {1/2/3,
        2/3/4,
        3/4/5,
        4/5/1,
        5/1/2}
        {
        \begin{scope}[xshift=200,yshift=-100*\shft]
            \pent
            \foreach \i in {1,...,5}
                \draw[threegreen] (p\j) -- (p\k) -- (p\l) -- cycle;
            \pent
        \end{scope}
        }
    
    % Fourth row
        
    \foreach \j/\k/\l [count=\shft from 0] in   
       {1/2/4,
        2/3/5,
        3/4/1,
        4/5/2,
        5/1/3}
        {
        \begin{scope}[xshift=300,yshift=-100*\shft]
            \pent
            \foreach \i in {1,...,5}
                \draw[threegreen] (p\j) -- (p\k) -- (p\l) -- cycle;
            \pent
        \end{scope}
        }
\end{tikzpicture}

\end{document}

Answer 3

类似这样的事情可能是一个起点：

\documentclass[tikz, border=2mm]{standalone}
\usetikzlibrary{shapes.geometric}

\begin{document}
\begin{tikzpicture}

\node[regular polygon, regular polygon sides=5, minimum size=2cm] (A){};
\filldraw[line width=3pt, green, line join=round] (A.corner 1)--(A.corner 2)--(A.corner 4)--(A.corner 5)--cycle;
\foreach \i in {1,2,...,5}
    \filldraw (A.corner \i) circle(1pt);

\end{tikzpicture}
\end{document}