1998 AIME 第 15 题

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15.

定义一张多米诺牌为一个由不同正整数组成的有序对。一个合法的多米诺序列是一列互不相同的多米诺牌,其中第一张之后每张牌的第一坐标都等于前一张牌的第二坐标,并且对于任意 iijj(i,j)(i, j)(j,i)(j, i) 不能同时出现。令 D40D_{40} 为所有坐标不大于 4040 的多米诺牌组成的集合。求使用 D40D_{40} 中的多米诺牌所能形成的最长合法多米诺序列的长度。

Define a domino to be an ordered pair of distinct positive integers. A proper sequence of dominos is a list of distinct dominos in which the first coordinate of each pair after the first equals the second coordinate of the immediately preceding pair, and in which (i,j)(i, j) and (j,i)(j, i) do not both appear for any ii and j.j. Let D40D_{40} be the set of all dominos whose coordinates are no larger than 40.40. Find the length of the longest proper sequence of dominos that can be formed using the dominos of D40.D_{40}.

答案:761
知识点:图论奇偶性
难度评级:3160
解答:

一张多米诺牌 (i,j)(i, j) 是顶点 1,,401, \ldots, 40 上完全图的一条有向边,而 (i,j)(i, j)(j,i)(j, i) 不能同时出现的规则意味着 (402)=780\binom{40}{2} = 780 条无向边每条至多使用一次。一个合法序列正是一条迹:不重复边的游走。在任何迹中,除两个端点外,每个顶点进入和离开的次数相等,所以它在已用边集合中的度为偶数。

完全图中每个顶点的度都是奇数 3939,所以在未用边集合中至少有 3838 个顶点必须是奇度顶点,而一个有 3838 个奇度顶点的图至少有 382=19\frac{38}{2} = 19 条边。因此最多能使用 78019=761780 - 19 = 761 张多米诺牌。

反过来,先放弃 1919 条互不相交的边 (3,4)(3,4)(5,6)(5,6)\ldots(39,40)(39,40)。剩下的图连通,且只有顶点 1122 为奇度顶点,所以存在一条遍历剩余全部 761761 条边的欧拉迹;按这条迹行进的方向给每条边定向,就得到长度为 761761 的合法序列。

A domino (i,j)(i, j) is an oriented edge of the complete graph on vertices 1,,40,1, \ldots, 40, and the rule that (i,j)(i, j) and (j,i)(j, i) cannot both appear means each of the (402)=780\binom{40}{2} = 780 edges is available at most once. A proper sequence is exactly a trail: a walk that repeats no edge. In any trail, every vertex other than the two endpoints is entered and left equally often, so it has even degree in the set of edges used.

In the complete graph every vertex has odd degree 39,39, so at least 3838 vertices must have odd degree in the set of unused edges, and a graph with 3838 odd-degree vertices has at least 382=19\frac{38}{2} = 19 edges. Hence at most 78019=761780 - 19 = 761 dominos can be used.

Conversely, set aside the 1919 disjoint edges (3,4),(3,4), (5,6),(5,6), ,\ldots, (39,40).(39,40). The remaining graph is connected and only vertices 11 and 22 have odd degree, so it has an Euler trail traversing all 761761 remaining edges; orienting each edge in the direction of travel gives a proper sequence of length 761.761.

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