#P1682C. LIS or Reverse LIS?

    ID: 260 远端评测题 1000ms 256MiB 尝试: 32 已通过: 6 难度: 8 上传者: 标签>constructive algorithmsgreedyimplementationmath*1400

LIS or Reverse LIS?

Description

You are given an array $a$ of $n$ positive integers.

Let $\text{LIS}(a)$ denote the length of longest strictly increasing subsequence of $a$. For example,

  • $\text{LIS}([2, \underline{1}, 1, \underline{3}])$ = $2$.
  • $\text{LIS}([\underline{3}, \underline{5}, \underline{10}, \underline{20}])$ = $4$.
  • $\text{LIS}([3, \underline{1}, \underline{2}, \underline{4}])$ = $3$.

We define array $a'$ as the array obtained after reversing the array $a$ i.e. $a' = [a_n, a_{n-1}, \ldots , a_1]$.

The beauty of array $a$ is defined as $min(\text{LIS}(a),\text{LIS}(a'))$.

Your task is to determine the maximum possible beauty of the array $a$ if you can rearrange the array $a$ arbitrarily.

题面翻译

设一个长为 nn 的整数序列 aa{a1,a2,a3,,an}\{a_1,a_2,a_3,\cdots,a_n\},那么 aa' 表示 {an,an1,an2,,a1}\{a_n,a_{n-1},a_{n-2},\cdots,a_1\}LIS(a)\operatorname{LIS}(a) 表示 aa 的最长严格上升子序列的长度。

现在给定 aa 数组,请你将 aa 数组重新排列,使得重排后的 min(LIS(a),LIS(a))\min(\operatorname{LIS}(a),\operatorname{LIS}(a')) 最大。

输入 tt 组数据,每组数据先输入 nn ,然后输入 nn 个整数,所有 nn 之和不超过 2×1052 \times 10^5

输出 tt 行,每行一组数据的答案,按输入顺序输出。

Input

The input consists of multiple test cases. The first line contains a single integer $t$ $(1 \leq t \leq 10^4)$  — the number of test cases. Description of the test cases follows.

The first line of each test case contains a single integer $n$ $(1 \leq n \leq 2\cdot 10^5)$  — the length of array $a$.

The second line of each test case contains $n$ integers $a_1,a_2, \ldots ,a_n$ $(1 \leq a_i \leq 10^9)$  — the elements of the array $a$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2\cdot 10^5$.

Output

For each test case, output a single integer  — the maximum possible beauty of $a$ after rearranging its elements arbitrarily.

3
3
6 6 6
6
2 5 4 5 2 4
4
1 3 2 2
1
3
2

Note

In the first test case, $a$ = $[6, 6, 6]$ and $a'$ = $[6, 6, 6]$. $\text{LIS}(a) = \text{LIS}(a')$ = $1$. Hence the beauty is $min(1, 1) = 1$.

In the second test case, $a$ can be rearranged to $[2, 5, 4, 5, 4, 2]$. Then $a'$ = $[2, 4, 5, 4, 5, 2]$. $\text{LIS}(a) = \text{LIS}(a') = 3$. Hence the beauty is $3$ and it can be shown that this is the maximum possible beauty.

In the third test case, $a$ can be rearranged to $[1, 2, 3, 2]$. Then $a'$ = $[2, 3, 2, 1]$. $\text{LIS}(a) = 3$, $\text{LIS}(a') = 2$. Hence the beauty is $min(3, 2) = 2$ and it can be shown that $2$ is the maximum possible beauty.