#P7945. 「Wdcfr-1」Yet Another Cirno Game (easy version)

「Wdcfr-1」Yet Another Cirno Game (easy version)

题目描述

The only difference between the two versions is whether you have to find a way to get the maximum points.

Cirno drew a graph. This graph contains 4n4\cdot n nodes, numbered 00 through 4n14\cdot n - 1. Also:

  • For any 0i30\le i\le 3 and 0j,k<n0 \le j, k \lt n, node (ni+j)(n\cdot i + j) and node (ni+k)(n\cdot i + k) are connected.
  • For any 0i<n0 \le i < n and 0j,k30 \le j, k \le 3, node (i+nj)(i + n\cdot j) and node (i+nk)(i + n\cdot k) are connected.

Cirno called Daiyousei to come and play with her.

The rules for this game are as follows:

  • Firstly, Cirno chooses 2n2\cdot n (i.e. half) of the nodes, and she colors them blue. The rest are left red.
  • Then there are 2n2\cdot n turns: for each turn Cirno first chooses a blue node, and Daiyousei chooses a red node. If those two nodes are connected, Daiyousei gets a point.

Try to maximize the number of points Daiyousei gets.

输入格式

The first line contains an integer nn. Then one line below, 2n2\cdot n numbers follow: they are the nodes that Cirno chose.

输出格式

For each test case, output an integer xx in a single line, which is the maximum number of points Daiyousei can get.

You don't have to output anything else in this version.

题目大意

两个难度之间的唯一区别为是否需要输出对应的匹配方案

给定 4n4n 个点,点的编号从 004n14n-1,每个点有两种属性 xxyy,编号为 ii 的点的属性为 $x_i=\left\lfloor\frac{i}{n}\right\rfloor,y_i=i-n\left\lfloor\frac{i}{n}\right\rfloor$。对于每两个点 i,j(0i<j<4n)i,j(0\le i <j < 4n) 之间,若 xi=xjx_i= x_jyi=yjy_i= y_j,则这两点有一条无向边相连。现有其中 2n2n 个不同的点被选择,试给每个被选择的点都匹配上另一个点,使得:

  • 被匹配到的点未被选择;
  • 每个点都只被匹配一次;
  • 若每个点与其匹配的点之间有连边,那么记为一分,需要最大化分值。

请输出最大的分值。

3
0 1 2 3 4 5
6

提示

Explanation

In the following picture, nodes in matrices are connected to each other. Cirno chose nodes 0,1,2,3,4,50,1,2,3,4,5.

Arrows below show a possible way for Daiyousei to get the maximum number of points she can get.

Constraints

1n2×1061\le n\le 2\times 10^6.