当前你的浏览器版本过低,网站已在兼容模式下运行,兼容模式仅提供最小功能支持,网站样式可能显示不正常。
请尽快升级浏览器以体验网站在线编辑、在线运行等功能。

建议使用的浏览器:

谷歌Chrome 火狐Firefox Opera浏览器 微软Edge浏览器 QQ浏览器 360浏览器 傲游浏览器

5087:Revenge of LIS II

题目描述
In computer science, the longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence's elements are in sorted order, lowest to highest, and in which the subsequence is as long as possible. This subsequence is not necessarily contiguous, or unique.
---Wikipedia

Today, LIS takes revenge on you, again. You mission is not calculating the length of longest increasing subsequence, but the length of the second longest increasing subsequence.
Two subsequence is different if and only they have different length, or have at least one different element index in the same place. And second longest increasing subsequence of sequence S indicates the second largest one while sorting all the increasing subsequences of S by its length.
输入解释
The first line contains a single integer T, indicating the number of test cases.

Each test case begins with an integer N, indicating the length of the sequence. Then N integer Ai follows, indicating the sequence.

[Technical Specification]
1. 1 <= T <= 100
2. 2 <= N <= 1000
3. 1 <= Ai <= 1 000 000 000
输出解释
For each test case, output the length of the second longest increasing subsequence.
输入样例
3
2
1 1
4
1 2 3 4
5
1 1 2 2 2
输出样例
1
3
2
提示
For the first sequence, there are two increasing subsequence: [1], [1]. So the length of the second longest increasing subsequence is also 1, same with the length of LIS.
来自杭电HDUOJ的附加信息
Recommend heyang

该题目是Virtual Judge题目,来自 杭电HDUOJ

题目来源 BestCoder Round #16

源链接: HDU-5087

最后修改于 2020-10-25T23:19:41+00:00 由爬虫自动更新

共提交 0

通过率 --%
时间上限 内存上限
2000/1000MS(Java/Others) 32768/32768K(Java/Others)