The only difference between easy and hard versions is constraints.
You are given a sequence aa consisting of nn positive integers.
Let's define a three blocks palindrome as the sequence, consisting of at most two distinct elements (let these elements are aa and bb, aa can be equal bb) and is as follows: [a,a,…,a������������x,b,b,…,b����������y,a,a,…,a������������x][a,a,…,a⏟x,b,b,…,b⏟y,a,a,…,a⏟x]. There x,yx,y are integers greater than or equal to 00. For example, sequences [][], [2][2], [1,1][1,1], [1,2,1][1,2,1], [1,2,2,1][1,2,2,1] and [1,1,2,1,1][1,1,2,1,1] are three block palindromes but [1,2,3,2,1][1,2,3,2,1], [1,2,1,2,1][1,2,1,2,1] and [1,2][1,2] are not.
Your task is to choose the maximum by length subsequence of aa that is a three blocks palindrome.
You have to answer tt independent test cases.
Recall that the sequence tt is a a subsequence of the sequence ss if tt can be derived from ss 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]s=[1,2,1,3,1,2,1], then possible subsequences are: [1,1,1,1][1,1,1,1], [3][3] and [1,2,1,3,1,2,1][1,2,1,3,1,2,1], but not [3,2,3][3,2,3] and [1,1,1,1,2][1,1,1,1,2].
The first line of the input contains one integer tt (1≤t≤1041≤t≤104) — the number of test cases. Then tt test cases follow.
The first line of the test case contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the length of aa. The second line of the test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤2001≤ai≤200), where aiai is the ii-th element of aa. Note that the maximum value of aiai can be up to 200200.
It is guaranteed that the sum of nn over all test cases does not exceed 2⋅1052⋅105 (∑n≤2⋅105∑n≤2⋅105).
For each test case, print the answer — the maximum possible length of some subsequence of aa that is a three blocks palindrome.
6 8 1 1 2 2 3 2 1 1 3 1 3 3 4 1 10 10 1 1 26 2 2 1 3 1 1 1
7 2 4 1 1 3
用1—200之间的数分别作两边，前缀和加暴力
用长整型开数组会翻车
#include <iostream> #include <vector> #include <algorithm> #include <string> #include <set> #include <queue> #include <map> #include <sstream> #include <cstdio> #include <cstring> #include <numeric> #include <cmath> #include <iomanip> #include <deque> #include <bitset> //#include <unordered_set> //#include <unordered_map> #define ll long long #define pii pair<int, int> #define rep(i,a,b) for(int i=a;i<=b;i++) #define dec(i,a,b) for(int i=a;i>=b;i--) #define forn(i, n) for(int i = 0; i < int(n); i++) using namespace std; int dir[4][2] = { { 1,0 },{ 0,1 } ,{ 0,-1 },{ -1,0 } }; const long long INF = 0x7f7f7f7f7f7f7f7f; const int inf = 0x3f3f3f3f; const double pi = 3.14159265358979323846; const double eps = 1e-6; const int mod = 1e9 + 7; const int N = 3e3 + 5; //if(x<0 || x>=r || y<0 || y>=c) inline ll read() { ll x = 0; bool f = true; char c = getchar(); while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); } while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar(); return f ? x : -x; } ll gcd(ll m, ll n) { return n == 0 ? m : gcd(n, m % n); } ll lcm(ll m, ll n) { return m * n / gcd(m, n); } bool prime(int x) { if (x < 2) return false; for (int i = 2; i * i <= x; ++i) { if (x % i == 0) return false; } return true; } ll qpow(ll m, ll k, ll mod) { ll res = 1, t = m; while (k) { if (k & 1) res = res * t % mod; t = t * t % mod; k >>= 1; } return res; } bool check(string s,string s1) { int cnt = 0; forn(i, s.size()) { if (s[i] != s1[i]) cnt++; } if (cnt > 1) return false; return true; } int main() { int T; cin >> T; while (T--) { int n; cin >> n; vector<int> a(n + 1); rep(i, 1, n) cin >> a[i]; vector<vector<int>> dp(n + 1, vector<int>(205)),pos(205); rep(i, 1, n) { dp[i] = dp[i - 1]; dp[i][a[i]]++; pos[a[i]].push_back(i); } int res = 1; rep(i, 1, 200) { forn(k, pos[i].size()/2) { int sz = pos[i].size(), y = 0; int l = pos[i][k], r = pos[i][sz - k - 1]-1; rep(j, 1, 200) { y = max(dp[r][j] - dp[l][j], y); } res = max(res, (k+1) * 2 + y); } } cout << res << endl; } return 0; }