제출 #763218

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
763218		penguin133	서열 (APIO23_sequence)	C++17	100 / 100	1704 ms	135228 KiB

sequence

#include <bits/stdc++.h>
using namespace std;
 
//#define int long long
#define pi pair<int, int>
#define pii pair<int, pi>
#define fi first
#define se second
#ifdef _WIN32
#define getchar_unlocked _getchar_nolock
#endif
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
int n;
struct node{
	int s, e, m;
	int mx, mn, lazy;
	node *l, *r;
	node(int _s, int _e){
		s = _s, e = _e, m = (s + e) >> 1;
		if(s != e)l = new node(s, m), r = new node(m+1, e);
		lazy = 0;
		mn = s + 1, mx = e + 1;
	}
	
	inline void prop(){
		if(!lazy)return;
		mx += lazy, mn += lazy;
		if(s != e)l->lazy += lazy, r->lazy += lazy;
		lazy = 0;
	}
	
	inline void upd(int a, int b, int c){
		if(a == s && b == e)lazy += c;
		else{
			if(b <= m)l->upd(a, b, c);
			else if(a > m)r->upd(a, b, c);
			else l->upd(a, m, c), r->upd(m+1, b, c);
			l->prop(), r->prop();
			mn= min(l->mn, r->mn);
			mx=  max(l->mx, r->mx);
		}
	}
	
	inline pi qry(int a, int b){
		prop();
		if(a == s && b == e)return {mn, mx};
		if(b <= m)return l->qry(a, b);
		else if(a > m)return r->qry(a, b);
		else{
			pi lft = l->qry(a, m), rgt = r->qry(m+1 ,b);
			return {min(lft.fi, rgt.fi), max(lft.se, rgt.se)};
		}
	}
	
	inline int qm(int a, int b){
		prop();
		if(a == s && b == e)return mx;
		if(b <= m)return l->qm(a, b);
		if(a > m)return r->qm(a, b);
		return max(l->qm(a, m), r->qm(m+1, b));
	}
	
	inline int qmm(int a, int b){
		prop();
		if(a == s && b == e)return mn;
		if(b <= m)return l->qmm(a, b);
		if(a > m)return r->qmm(a, b);
		return min(l->qmm(a, m), r->qmm(m+1, b));
	}
	
}*lft;
 
struct node2{
	int s, e, m;
	int mx, mn, lazy;
	node2 *l, *r;
	node2(int _s, int _e){
		s = _s, e = _e, m = (s + e) >> 1;
		if(s != e)l = new node2(s, m), r = new node2(m+1, e);
		lazy = 0, mx = (n - s), mn = (n - e);
	}
	
	inline void prop(){
		if(!lazy)return;
		mx += lazy, mn += lazy;
		if(s != e)l->lazy += lazy, r->lazy += lazy;
		lazy = 0;
	}
	
	inline void upd(int a, int b, int c){
		if(a == s && b == e)lazy += c;
		else{
			if(b <= m)l->upd(a, b, c);
			else if(a > m)r->upd(a, b, c);
			else l->upd(a, m, c), r->upd(m+1, b, c);
			l->prop(), r->prop();
			mn= min(l->mn, r->mn);
			mx=  max(l->mx, r->mx);
		}
	}
	inline int qm(int a, int b){
		prop();
		if(a == s && b == e)return mx;
		if(b <= m)return l->qm(a, b);
		if(a > m)return r->qm(a, b);
		return max(l->qm(a, m), r->qm(m+1, b));
	}
	
	inline int qmm(int a, int b){
		prop();
		if(a == s && b == e)return mn;
		if(b <= m)return l->qmm(a, b);
		if(a > m)return r->qmm(a, b);
		return min(l->qmm(a, m), r->qmm(m+1, b));
	}
	
}*rgt;
 
vector <int> occ[500005];
int B[10][500001], lf[500005], rg[500005];
 
inline void upd(int f[], int l, int r, int v){
	l++;
	r += 2;
	for(;l<=n;l+=(l & -l))f[l] += v;
	for(;r<=n;r+=(r & -r))f[r] -= v;
}
 
inline int qr(int f[], int p){
	p++;
	int res = 0;
	for(;p;p-=(p & -p))res += f[p];
	return res;
}
 
int sequence(int N, vector <int> A){
	n = N;
	for(int i=0;i<N;i++)occ[A[i]].push_back(i);
	lft = new node(0, N - 1);
	rgt = new node2(0, N - 1);
	int ans = 1;
	for(int i=0;i<N;i++){
		upd(lf, i, N, 1);
		upd(rg, 0, i, 1);
	}
	for(int i=1;i<=N;i++){
		
		int in = 0;
		for(int j = 0; j < (int)occ[i].size(); j++){
			int r = occ[i][j];
			int tmp2 = lft->qm(r, N - 1), ref2 = qr(lf, r), tmp = rgt->qm(0, r);
			int ref = qr(rg, r);
			tmp -= ref;
			int tmp5 = (r == 0 ? 0 : ref2 - 1);
          
			tmp2 -= ref2; 
			B[0][j] = tmp, B[1][j] = ref, B[2][j] = tmp2, B[3][j] = ref2, B[8][j] = tmp5;
		}
		
		for(auto j : occ[i]){
			lft->upd(j, N - 1, -2);
			rgt->upd(0, j, -2);
			upd(lf, j, N - 1, -2);
			upd(rg, 0, j, -2);
		}
		for(int j = 0; j < (int)occ[i].size(); j++){
			int r = occ[i][j];
			int tmp4 = lft->qmm(r, N - 1), ref4 = qr(lf, r);
			tmp4 -= ref4;
			
			int tmp3 = rgt->qmm(0, r);
			int ref3 = qr(rg, r);
			
			tmp3 -= ref3;
			int tmp6 = (r == 0 ? 0 : ref4 + 1);
			B[4][j] = tmp3, B[5][j] = ref3, B[6][j] = tmp4, B[7][j] = ref4, B[9][j] = tmp6;
			int tmp2 = B[2][j], ref2 = B[3][j];
			while(in + ans <= j){
				int tmp = B[0][in];
				tmp3 = B[4][in], ref3 = B[5][in];
				tmp += tmp2;
				tmp3 += tmp4;
				int val = ref2 - B[8][in];
				int val2 = ref4 - B[9][in];
				long long mn = val + tmp, mx = val2 + tmp3;
				if(mn * mx > 0)in++;
				else break;
			}
			//cout << i << ' ' << j << ' ' << in << '\n';
			ans = max(ans, j - in + 1);
        }
	}
	return ans;
}

• Subtask 1: 11 / 11
• Subtask 2: 17 / 17
• Subtask 3: 7 / 7
• Subtask 4: 12 / 12
• Subtask 5: 13 / 13
• Subtask 6: 22 / 22
• Subtask 7: 18 / 18