E

The only difference between easy and hard versions is constraints.

You are given a sequence a consisting of n positive integers.

Let's define a three blocks palindrome as the sequence, consisting of at most two distinct elements (let these elements are a and b, a can be equal b) and is as follows: [a,a,…,a������������x,b,b,…,b����������y,a,a,…,a������������x]. There x,y are integers greater than or equal to 0. For example, sequences [], [2], [1,1], [1,2,1], [1,2,2,1] and [1,1,2,1,1] are three block palindromes but [1,2,3,2,1], [1,2,1,2,1] and [1,2] are not.

Your task is to choose the maximum by length subsequence of a that is a three blocks palindrome.

You have to answer t independent test cases.

Recall that the sequence t is a a subsequence of the sequence s if t can be derived from s by removing zero or more elements without changing the order of the remaining elements. For example, if s=[1,2,1,3,1,2,1], then possible subsequences are: [1,1,1,1], [3] and [1,2,1,3,1,2,1], but not [3,2,3] and [1,1,1,1,2].

题意杀我，一定要明白是找到aaabbbaaa这样的回文串，不是普通的回文串，而且子序列的相对位置不可以变，可是明白了题意之后我满脑子都是2^n的算法，显然是不现实的。借用sc学长的题解做法：我们使用三重循环：第一重枚举a是什么，第二重枚举取多少个a，第三种枚举b取什么。这样的话，我们知道，取同样多的a，中间的区间越大，b的个数可能就越多，所以我们取前x个a和后x个a，在这中间找b，由此就有了一个优化：我们记录每一个数出现的所有位置，也记录每一个数在每一个位置出现的前缀和。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
using namespace std;
const int maxn=2e3+5;
int t;
int n;
int a[maxn];
int ans;
int s[maxn][30];
vector<int>v[30];
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        memset(s,0,sizeof(s));
        ans=0;
        for(int i=0;i<30;++i)v[i].clear();
        for(int i=1;i<=n;++i)
        {
            scanf("%d",&a[i]),v[a[i]].push_back(i);
            for(int j=1;j<=26;++j)
                s[i][j]=s[i-1][j];
            s[i][a[i]]++;//前缀和 
        }
        for(int i=1;i<=26;++i)
        {
            ans=max(ans,s[n][i]);
            int hal=(s[n][i]>>1);
            for(int j=1;j<=hal;++j)//枚举取多少个a 
            {
                int l=v[i][j-1]+1,r=v[i][s[n][i]-j]-1;
                if(r>=l)
                {
                    for(int k=1;k<=26;++k)
                    {
                        ans=max(ans,2*j+s[r][k]-s[l-1][k]);
                    }
                }
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}