这个图是由一个程序生成的，我没有输入这个（我不希望任何人想到我会这样，这种格式是如此......人为的）
digraph G {
rankdir=LR
n0_0_ [label = "[0, 0]"]
n0_0_ -> n0_0_
n0_1_ [label = "[0, 1]"]
n0_0_ -> n0_1_
n0_1_ -> n0_0_
n1_0_ [label = "[1, 0]"]
n0_0_ -> n1_0_
n1_0_ -> n0_0_
n1_1_ [label = "[1, 1]"]
n0_0_ -> n1_1_
n1_1_ -> n0_0_
}
Graphviz在左侧放置了[0,0]，在右侧放置了下一个等级。
然而，在中心和其余3个节点周围有[0,0]更有意义。
从[0,0]到它自身的弧将被省略，对于应用程序来说重要的是graphviz没有尝试用等级构造它，可以开发可爱的对称性，为什么我不希望graphviz看到它和利用它，它使用的方法（弹簧IIRC上的能量方法）应该创造正确的形式！
如何说“没有rankdir”或其他相似之处？是有向图的副产品吗？
修改
不，它不是有向图的副产品，正如你可以看到的那样，我想另一种方式来说这个目标是“没有交叉边缘”，图形是平面的。
graph G {
n0_0_ [label = "[0, 0]"]
n0_0_ -- n0_0_
n0_1_ [label = "[0, 1]"]
n0_0_ -- n0_1_
n0_1_ -- n0_2_
n0_2_ -- n0_3_
n0_3_ -- n0_4_
n0_4_ -- n0_0_
n0_2_ [label = "[0, 2]"]
n0_0_ -- n0_2_
n0_2_ -- n0_4_
n0_4_ -- n0_1_
n0_1_ -- n0_3_
n0_3_ -- n0_0_
n0_3_ [label = "[0, 3]"]
n0_0_ -- n0_3_
n0_3_ -- n0_1_
n0_1_ -- n0_4_
n0_4_ -- n0_2_
n0_2_ -- n0_0_
n0_4_ [label = "[0, 4]"]
n0_0_ -- n0_4_
n0_4_ -- n0_3_
n0_3_ -- n0_2_
n0_2_ -- n0_1_
n0_1_ -- n0_0_
n1_0_ [label = "[1, 0]"]
n0_0_ -- n1_0_
n1_0_ -- n2_0_
n2_0_ -- n3_0_
n3_0_ -- n4_0_
n4_0_ -- n0_0_
n1_1_ [label = "[1, 1]"]
n0_0_ -- n1_1_
n1_1_ -- n2_2_
n2_2_ -- n3_3_
n3_3_ -- n4_4_
n4_4_ -- n0_0_
n1_2_ [label = "[1, 2]"]
n0_0_ -- n1_2_
n1_2_ -- n2_4_
n2_4_ -- n3_1_
n3_1_ -- n4_3_
n4_3_ -- n0_0_
n1_3_ [label = "[1, 3]"]
n0_0_ -- n1_3_
n1_3_ -- n2_1_
n2_1_ -- n3_4_
n3_4_ -- n4_2_
n4_2_ -- n0_0_
n1_4_ [label = "[1, 4]"]
n0_0_ -- n1_4_
n1_4_ -- n2_3_
n2_3_ -- n3_2_
n3_2_ -- n4_1_
n4_1_ -- n0_0_
n2_0_ [label = "[2, 0]"]
n0_0_ -- n2_0_
n2_0_ -- n4_0_
n4_0_ -- n1_0_
n1_0_ -- n3_0_
n3_0_ -- n0_0_
n2_1_ [label = "[2, 1]"]
n0_0_ -- n2_1_
n2_1_ -- n4_2_
n4_2_ -- n1_3_
n1_3_ -- n3_4_
n3_4_ -- n0_0_
n2_2_ [label = "[2, 2]"]
n0_0_ -- n2_2_
n2_2_ -- n4_4_
n4_4_ -- n1_1_
n1_1_ -- n3_3_
n3_3_ -- n0_0_
n2_3_ [label = "[2, 3]"]
n0_0_ -- n2_3_
n2_3_ -- n4_1_
n4_1_ -- n1_4_
n1_4_ -- n3_2_
n3_2_ -- n0_0_
n2_4_ [label = "[2, 4]"]
n0_0_ -- n2_4_
n2_4_ -- n4_3_
n4_3_ -- n1_2_
n1_2_ -- n3_1_
n3_1_ -- n0_0_
n3_0_ [label = "[3, 0]"]
n0_0_ -- n3_0_
n3_0_ -- n1_0_
n1_0_ -- n4_0_
n4_0_ -- n2_0_
n2_0_ -- n0_0_
n3_1_ [label = "[3, 1]"]
n0_0_ -- n3_1_
n3_1_ -- n1_2_
n1_2_ -- n4_3_
n4_3_ -- n2_4_
n2_4_ -- n0_0_
n3_2_ [label = "[3, 2]"]
n0_0_ -- n3_2_
n3_2_ -- n1_4_
n1_4_ -- n4_1_
n4_1_ -- n2_3_
n2_3_ -- n0_0_
n3_3_ [label = "[3, 3]"]
n0_0_ -- n3_3_
n3_3_ -- n1_1_
n1_1_ -- n4_4_
n4_4_ -- n2_2_
n2_2_ -- n0_0_
n3_4_ [label = "[3, 4]"]
n0_0_ -- n3_4_
n3_4_ -- n1_3_
n1_3_ -- n4_2_
n4_2_ -- n2_1_
n2_1_ -- n0_0_
n4_0_ [label = "[4, 0]"]
n0_0_ -- n4_0_
n4_0_ -- n3_0_
n3_0_ -- n2_0_
n2_0_ -- n1_0_
n1_0_ -- n0_0_
n4_1_ [label = "[4, 1]"]
n0_0_ -- n4_1_
n4_1_ -- n3_2_
n3_2_ -- n2_3_
n2_3_ -- n1_4_
n1_4_ -- n0_0_
n4_2_ [label = "[4, 2]"]
n0_0_ -- n4_2_
n4_2_ -- n3_4_
n3_4_ -- n2_1_
n2_1_ -- n1_3_
n1_3_ -- n0_0_
n4_3_ [label = "[4, 3]"]
n0_0_ -- n4_3_
n4_3_ -- n3_1_
n3_1_ -- n2_4_
n2_4_ -- n1_2_
n1_2_ -- n0_0_
n4_4_ [label = "[4, 4]"]
n0_0_ -- n4_4_
n4_4_ -- n3_3_
n3_3_ -- n2_2_
n2_2_ -- n1_1_
n1_1_ -- n0_0_
}
它真的不能很好地表达这一点。
ordering=out有所帮助，但仍然有非常明确的排名
答案 0 :(得分：1)
dot应该适用于有向图并使用节点排名。
neato或circo或sfdp等其他布局可能会在您的情况下产生更好的结果。