AI 中文总结
研究判定树与欧几里得最小生成树接近程度的问题，刻画了有EMST图的毛毛虫树，给出线性时间判定算法及计算图的方法，还对特定树给出坏对数量上界并构造特殊图。
AI 中文摘要
设$\Gamma$是树$T$的无交叉直线图。$\Gamma$中的“坏对”是$T$中一对不相邻顶点，其在$\Gamma$中的欧几里得距离小于在$\Gamma$中连接它们的路径上最长边的长度。当$\Gamma$没有坏对时，$\Gamma$是其顶点集的欧几里得最小生成树（简称EMST图）。已知判定最大度至多为六的树是否有EMST图是NP难的。相比之下，我们刻画了那些有EMST图的毛毛虫树。该刻画产生了一个线性时间算法，可判定毛毛虫树是否有EMST图，若是则计算出这样的图。对于最大度为六的毛毛虫树，我们还给出一个线性时间算法来计算具有最少坏对的无交叉直线图。对于最大顶点度为$\Delta$的$n$顶点树，我们证明了坏对最小数量的$\Delta^2n\log n$上界。在星型树的特殊情况下，我们构造了具有最少坏对的图。
英文摘要
Let $Γ$ be a straight-line crossing-free drawing of a tree $T$. A \emph{bad pair} in $Γ$ is a pair of non-adjacent vertices of $T$ whose Euclidean distance in $Γ$ is smaller than the length of the longest edge in the path connecting them in~$Γ$. When $Γ$ has no bad pairs, $Γ$ is a Euclidean Minimum Spanning Tree of its vertex set (or EMST-drawing for short). Deciding whether a tree of maximum degree at most six admits an EMST-drawing is known to be \NP-hard. In contrast, we characterize those caterpillars that admit an EMST-drawing. The characterization gives rise to a linear-time algorithm that decides if a caterpillar admits an EMST-drawing, and in the affirmative case, computes such a drawing. For caterpillars of maximum degree six, we further present a linear-time algorithm to compute a crossing-free straight-line drawing with the minimum number of bad pairs. For $n$-vertex trees with maximum vertex degree $Δ$, we prove the $Δ^2n\log n$ upper bound on the minimum number of bad pairs. In the special case of stars, we construct a drawing with the minimum number of bad pairs.
CommentsExtended version of the paper accepted to "34th International Symposium on Graph Drawing and Network Visualization" (GD 2026)
