2021 AMC 12A Fall 第 16 题

先试着解答 2021 AMC 12A Fall 第 16 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2021 AMC 12A Fall 解答,或核对答案

所有题目均经美国数学协会(MAA)官方合法授权使用。

16.

一个组织有 3030 名员工,其中 2020 人使用 A 品牌电脑,另外 1010 人使用 B 品牌电脑。出于安全原因,电脑只能彼此相连,且只能用电缆连接。电缆只能把一台 A 品牌电脑和一台 B 品牌电脑相连。如果两名员工的电脑直接由电缆相连,或可以通过一系列相连的电脑转发消息,则他们可以相互通信。最初没有任何电脑与其他电脑相连。一名技术员任意选择一台每种品牌的电脑,并在它们之间安装电缆,前提是这对电脑之间还没有电缆。技术员在每名员工都能相互通信时停止。最多可能使用多少根电缆?

An organization has 3030 employees, 2020 of whom have a brand A computer while the other 1010 have a brand B computer. For security, the computers can only be connected to each other and only by cables. The cables can only connect a brand A computer to a brand B computer. Employees can communicate with each other if their computers are directly connected by a cable or by relaying messages through a series of connected computers. Initially, no computer is connected to any other. A technician arbitrarily selects one computer of each brand and installs a cable between them, provided there is not already a cable between that pair. The technician stops once every employee can communicate with each other. What is the maximum possible number of cables used?

190190

191191

192192

195195

196196

答案:B
知识点:图论极端原理
难度评级:1840
小提示:

为了尽量延迟连通,尽可能久地保持图不连通,然后最后一根电缆把所有部分连接起来

To delay connectivity, keep the graph disconnected as long as possible, then one final cable joins everything

大提示:

在仍不连通的情况下,最多边数来自 1919 台 A 品牌和 1010 台 B 品牌形成的完全二分图,另留一台 A 品牌电脑孤立

The most edges while still disconnected is a complete bipartite graph on 1919 brand A and 1010 brand B, leaving one brand A computer isolated

解答:

技术员不断加电缆,直到图变为连通。为了最大化电缆数,要让网络尽可能久地保持不连通:留下一台 A 品牌电脑孤立,并把其余 1919 台 A 品牌电脑与全部 1010 台 B 品牌电脑完全相连。

这样在仍不连通时使用了 1910=19019 \cdot 10 = 190 根电缆。下一根电缆连接最后一台 A 品牌电脑,使所有人连通,总数为 190+1=191190 + 1 = 191

所以正确答案是 B

The technician keeps adding cables until the graph becomes connected. To maximize the count, keep the network disconnected for as long as possible: leave a single brand A computer isolated and fully connect the remaining 1919 brand A computers to all 1010 brand B computers.

That uses 1910=19019 \cdot 10 = 190 cables while still disconnected. The next cable connects the last brand A computer, joining everyone, for a total of 190+1=191.190 + 1 = 191.

Thus, the correct answer is B.

第 15 题#15
完整试卷

其他年份的第 16 题

1950 AMC 12 · 1951 AMC 12 · 1952 AMC 12 · 1953 AMC 12 · 1954 AMC 12 · 1955 AMC 12 · 1956 AMC 12 · 1957 AMC 12 · 1958 AMC 12 · 1959 AMC 12 · 1960 AMC 12 · 1961 AMC 12 · 1962 AMC 12 · 1963 AMC 12 · 1964 AMC 12 · 1965 AMC 12 · 1966 AMC 12 · 1967 AMC 12 · 1968 AMC 12 · 1969 AMC 12 · 1970 AMC 12 · 1971 AMC 12 · 1972 AMC 12 · 1973 AMC 12 · 1974 AMC 12 · 1975 AMC 12 · 1976 AMC 12 · 1977 AMC 12 · 1978 AMC 12 · 1979 AMC 12 · 1980 AMC 12 · 1981 AMC 12 · 1982 AMC 12 · 1983 AMC 12 · 1984 AMC 12 · 1985 AMC 12 · 1986 AMC 12 · 1987 AMC 12 · 1988 AMC 12 · 1989 AMC 12 · 1990 AMC 12 · 1991 AMC 12 · 1992 AMC 12 · 1993 AMC 12 · 1994 AMC 12 · 1995 AMC 12 · 1996 AMC 12 · 1997 AMC 12 · 1998 AMC 12 · 1999 AMC 12 · 2000 AMC 12 · 2001 AMC 12 · 2002 AMC 12A · 2002 AMC 12B · 2003 AMC 12A · 2003 AMC 12B · 2004 AMC 12A · 2004 AMC 12B · 2005 AMC 12A · 2005 AMC 12B · 2006 AMC 12A · 2006 AMC 12B · 2007 AMC 12A · 2007 AMC 12B · 2008 AMC 12A · 2008 AMC 12B · 2009 AMC 12A · 2009 AMC 12B · 2010 AMC 12A · 2010 AMC 12B · 2011 AMC 12A · 2011 AMC 12B · 2012 AMC 12A · 2012 AMC 12B · 2013 AMC 12A · 2013 AMC 12B · 2014 AMC 12A · 2014 AMC 12B · 2015 AMC 12A · 2015 AMC 12B · 2016 AMC 12A · 2016 AMC 12B · 2017 AMC 12A · 2017 AMC 12B · 2018 AMC 12A · 2018 AMC 12B · 2019 AMC 12A · 2019 AMC 12B · 2020 AMC 12A · 2020 AMC 12B · 2021 AMC 12A Spring · 2021 AMC 12B Spring · 2021 AMC 12B Fall · 2022 AMC 12A · 2022 AMC 12B · 2023 AMC 12A · 2023 AMC 12B · 2024 AMC 12A · 2024 AMC 12B · 2025 AMC 12A · 2025 AMC 12B