【发布时间】:2018-12-15 01:24:08
【问题描述】:
我想用graphviz生成结构树,这样的从左到右的结构(请忽略颜色):
我得出的结论是我需要使用不可见的节点来实现这一点,我当前的代码如下所示:
digraph G {
node [ shape="box", width = 2, height = 1, fixedsize=true];
edge [arrowhead=none];
nodesep = 1;
ranksep=0.05;
splines = ortho;
rankdir = LR;
A1 [ shape="box", width = 2, height = 1, fixedsize=true];
B1 [ shape="box", width = 2, height = 1, fixedsize=true];
B2 [ shape="box", width = 2, height = 1, fixedsize=true];
B3 [ shape="box", width = 2, height = 1, fixedsize=true];
B4 [ shape="box", width = 2, height = 1, fixedsize=true];
B5 [ shape="box", width = 2, height = 1, fixedsize=true];
B6 [ shape="box", width = 2, height = 1, fixedsize=true];
W0 [ shape="circle", width = 0, height = 0, fixedsize=true, label=""];
W1 [ shape="circle", width = 0, height = 0, fixedsize=true, label=""];
W2 [ shape="circle", width = 0, height = 0, fixedsize=true, label=""];
W3 [ shape="circle", width = 0, height = 0, fixedsize=true, label=""];
W4 [ shape="circle", width = 0, height = 0, fixedsize=true, label=""];
W5 [ shape="circle", width = 0, height = 0, fixedsize=true, label=""];
W6 [ shape="circle", width = 0, height = 0, fixedsize=true, label=""];
subgraph {
A1 -> W0
W0 -> W3
W3 -> W2
W2 -> W1
W0 -> W4
W4 -> W5
W5 -> W6
W1 -> B1
W2 -> B2
W3 -> B3
W4 -> B4
W5 -> B5
W6 -> B6
{rank = same; A1;}
{rank = same; B1; B2; B3; B4; B5; B6;}
{rank = same; W0; W1; W2; W3; W4; W5; W6;}
}
}
使用点引擎我得到:
我的问题:
我可以强制某些节点占据中心位置(A1)吗?
我可以强制边缘连接到节点形状边框上的某些位置(例如:左中)吗?
也许有更好的方法来实现这种具有节点中心的树结构(我必须考虑到树的下一层可能相当复杂)
【问题讨论】:
标签: graphviz