제출 #1038096

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1038096		pan	서열 (APIO23_sequence)	C++17	0 / 100	661 ms	94288 KiB

sequence

#include <bits/stdc++.h>
//#include "bits_stdc++.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<ll, pi> pii;
typedef pair<pi, pi> piii;
 
// Condition for median: -c <= h - l <= c
 
struct node
{
 
    int s, e, m; 
    ll minn, maxx; 
    ll lazy; 
    node *l, *r; 
	node (int S, int E)
	{ 
	   s = S, e = E, m = (s+e)/2;
       minn = s, maxx = e;
       lazy = 0; 
       l=r=nullptr;
 
    }
    void create()
    {
	   if(s != e && l==nullptr)
       { 
           l = new node(s, m); 
           r = new node(m+1, e); 
       }
    }
 
	void propogate()
	{
       if (lazy==0) return; 
       create();
       minn+=lazy; 
       maxx += lazy;
       if (s != e){ 
           l->lazy+=lazy;
           r->lazy+=lazy;
       }
       lazy=0; 
    }
	void update(int S, int E, ll V)
	{ 
	   propogate();
       if(s==S && e==E) lazy += V; 
       else{ 
           create();
           if(E <= m) l->update(S, E, V); 
           else if (m < S) r->update(S, E, V); 
           else l->update(S, m, V),r->update(m+1, E, V);
           l->propogate(),r->propogate();
           maxx = max(l->maxx, r->maxx); 
           minn = min(l->minn, r->minn);
       }
    }
	pi query(int S, int E)
	{
       if (S> E) return mp(0LL, 0LL);
       propogate(); 
       if(s == S && e == E) return mp(minn, maxx);
       create();
       if(E <= m) return l->query(S, E); 
       else if(S >= m+1) return r->query(S, E); 
       else 
       {
		   pi a = l->query(S, m), b = r->query(m+1, E); 
		   return mp(min(a.f, b.f), max(a.s, b.s));
	   }
    }
 
} *root;
 
 
vector<ll> dis[500005];
 
 
int sequence(int N, std::vector<int> A)
{
	ll n = N;
	ll ans = 0;
	for (ll i=0; i<n; ++i) dis[A[i]].pb(i+1);
	root = new node(0, n);
	for (ll i=1; i<=n; ++i)
	{
		for (ll u: dis[i]) root->update(u, n, -1);
		ll N = dis[i].size();
		ll r = 0;
		for (ll l=0; l<N; ++l)
		{
			ll st = ((l==0)?0:dis[i][l-1]);
			while (r < N)
			{
				ll ee = ((r==N-1)?n:dis[i][r+1]-1);
				ll len = r-l+1;
				pi l1 = root->query(st, dis[i][l]-1), r1 = root->query(dis[i][r], ee);
				ll maxx = r1.s - l1.f;
				ll minn = r1.f - l1.s;
				if (maxx <= len && minn >= -len) {ans = max(ans, len); r++;}
				else break;
			}
		}
		for (ll u: dis[i]) root->update(u, n, -1);
	}
	return ans;
	
}

• Subtask 1: 0 / 11
• Subtask 2: 0 / 17
• Subtask 3: 0 / 7
• Subtask 4: 0 / 12
• Subtask 5: 0 / 13
• Subtask 6: 0 / 22
• Subtask 7: 0 / 18