For the game of Cops and Robbers, it is known that in 1-cop-win graphs, the cop can capture the robber in $\mathcal{O}(n)$ time, and that there exist graphs in which this capture time is tight. When $k\ge 2$, a simple counting argument shows that in k-cop-win graphs, the capture time is at most $\mathcal{O}(n^{k+1})$, however, no non-trivial lower bounds were previously known; indeed, in their 2011 book, Bonato and Nowakowski ask whether this upper bound can be improved. In this paper, the question of Bonato and Nowakowski is answered on the negative, proving that the $\mathcal{O}(n^{k+1})$ bound is asymptotically tight for any constant $k\ge 2$. This yields a surprising gap in the capture time complexities between the 1-cop and the 2-cop cases.

对于《警察与强盗》的游戏，众所周知，在1-cop-win图表中，警察可以捕获强盗$\mathcal{Ø}（ñ）$时间，并且存在其中捕获时间很短的图。什么时候$ķ\ge 2$，一个简单的计数参数表明，在k -cop-win图中，捕获时间最多为$\mathcal{Ø}（{ñ}^{ķ+\mathrm{1个}}）$但是，以前没有非平凡的下界。实际上，Bonato和Nowakowski在2011年的书中提出，是否可以提高这一上限。在本文中，Bonato和Nowakowski的问题得到了否定的回答，证明了$\mathcal{Ø}（{ñ}^{ķ+\mathrm{1个}}）$ 边界对于任何常数都是渐近紧的 $ķ\ge 2$。这在1-cop和2-cop病例之间的捕获时间复杂度上产生了令人惊讶的差距。