我如何获取这些到节点的特定路径？

Question

我有四个节点，一个父节点和三个子节点。我尝试用以下特定箭头路径进行链接：

我刚刚开始使用 tikz 包，到目前为止我的代码如下所示：

\usepackage{tikz}
\usetikzlibrary{arrows,positioning,calc}

\begin{document}

\begin{tikzpicture}[
,node distance =5mm
,>=latex'
,block/.style ={
    ,draw
    ,minimum height = 10mm
    ,minimum width =28mm
    ,align=center
    }
,every path/.style={->} %sets arrow style
]
\node [block]   (main) {Main};
\node [block, below=of main] (second)   {Second};  
\node [block, left=of second] (first)   {First};
\node [block, right=of second] (third)  {Third};

\draw (main) to (first);
\draw (main) to (second);
\draw (main) to (third);
\end{tikzpicture}

\end{document}

这给了我：

我尝试使用 -|，但只生成了路径中的一个角，而不是我需要的两个角。任何帮助都非常感谢。

Answer 1

像这样：

\begin{tikzpicture}[
,node distance =5mm
,>=latex'
,block/.style ={
    ,draw
    ,minimum height = 10mm
    ,minimum width =28mm
    ,align=center
    }
,every path/.style={->} %sets arrow style
]
\node [block]   (main) {Main};
\node [block, below=of main] (second)   {Second};
\node [block, left=of second] (first)   {First};
\node [block, right=of second] (third)  {Third};

\draw ([xshift=-3mm] main.south) -- + (0,-2mm) -| (first);
\draw (main) to (second);
\draw ([xshift= 3mm] main.south) -- + (0,-2mm) -| (third);
\end{tikzpicture}