| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 763213 | penguin133 | 서열 (APIO23_sequence) | C++17 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
//#define int long long
#include "sequence_apio23.h"
#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];
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;
}