Given a set $\mathcal{P}$ of points in the Euclidean plane and two triangulations of $\mathcal{P}$, the flip distance between these two triangulations is the minimum number of flips required to transform one triangulation into the other. The Parameterized Flip Distance problem is to decide if the flip distance between two given triangulations is equal to a given integer k. The previous best FPT algorithm runs in time $O^{⁎}(k\cdot c^k)$ ($c\le 2\times {14}^{11}$), where each step has fourteen possible choices, and the length of the action sequence is bounded by 11k. By analyzing the underlying properties of the flip sequence, each step of our algorithm has only five possible choices. Based on an auxiliary graph G, we prove that the length of the action sequence for our algorithm is bounded by $2|G|$. As a result, we present an FPT algorithm running in time $O^{⁎}(k\cdot {32}^k)$.

给定一组 $\mathcal{磷}$ 欧几里得平面上的点和两个三角剖分 $\mathcal{磷}$，这两个三角剖分之间的翻转距离是将一个三角剖分转换为另一个三角剖分所需的最少翻转次数。所述参数化距离翻转的问题是，以决定是否两个给定的三角之间的翻转距离等于一个给定的整数ķ。之前最好的FPT算法运行及时${哦}^{⁎}(克\cdot C^{克})$ ($C\le 2\times {14}^{11}$)，其中每个步骤有 14 个可能的选择，动作序列的长度以 11 k为界。通过分析翻转序列的基本属性，我们算法的每一步只有五种可能的选择。基于辅助图G，我们证明了我们算法的动作序列的长度为$2|G|$. 因此，我们提出了一个及时运行的 FPT 算法${哦}^{⁎}(克\cdot {32}^{克})$.