HDU 5078 Revenge of LIS II（dp LIS）_hdu 4996 revenge of lis-优快云博客

本文链接：https://blog.youkuaiyun.com/u014737310/article/details/40867627

本文介绍了一个计算机科学中的经典问题——寻找给定序列的第二长递增子序列。不同于最长递增子序列（LIS）问题，此问题需要找出所有递增子序列中长度次于最长的那个。文章提供了算法思路及实现代码，适用于算法竞赛和技术面试等场景。

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Problem Description

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.

Input

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

Output

For each test case, output the length of the second longest increasing subsequence.

Sample Input

Sample Output

  
   1
3
2

   
    
     Hint
    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.

Source

BestCoder Round #16

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参考链接：http://blog.youkuaiyun.com/acvay/article/details/40686171

比赛时没有读懂题目开始做结果被hack了，后来一直想nlogn的方法，无果，以后应该会想出来，以后再贴那种方法代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
using namespace std;
#define N 10005

int a[N],dp[N],c[N];
int n;

int main()
{
	int i,j,t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(i=1;i<=n;i++)
			scanf("%d",&a[i]);

        int ans=0;
        for(i=1;i<=n;i++)
		{
			dp[i]=1;
			c[i]=1;
			for(j=1;j<i;j++)
			{
			   if(a[i]<=a[j]) continue;

			   if(dp[j]+1>dp[i])
			   {
			   	 dp[i]=dp[j]+1;
			   	 c[i]=c[j];
			   }
			   else
				if(dp[j]+1==dp[i])
				 c[i]=2;
			}
			if(dp[i]>ans)
				ans=dp[i];
		}
		j=0;
		for(i=1;i<=n;i++)
			if(dp[i]==ans)
		{
			j+=c[i];
			if(j>1)
				break;
		}
		if(j>1)
			printf("%d\n",ans);
		else
			printf("%d\n",ans-1);
	}
	return 0;
}