제출 #1227087

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1227087		Lemser	Rectangles (IOI19_rect)	C++20	100 / 100	2478 ms	670908 KiB

rect

#include "rect.h"

#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx2")
#pragma GCC target("popcnt")
using namespace std;
 
using ll = long long;
using ull = unsigned long long;
using lld = long double;
using ii = pair<int,int>;
using pll = pair<ll, ll>;
 
using vi = vector<int>;
using vll = vector<ll>;
using vii = vector<ii>;
using vpll = vector<pll>;
using vlld = vector<lld>;
 
#define all(x) x.begin(),x.end()
#define lsb(x) x&(-x)
#define gcd(a,b) __gcd(a,b)
#define sz(x) (int)x.size()
#define pb push_back
#define fi first
#define se second
#define fls cout.flush()
 
#define fore(i, l, r) for (auto i = l; i < r; i++)
#define fo(i, n) fore (i, 0, n)
#define forex(i, r, l) for (auto i = r-1; i >= l; i--)
#define ffo(i, n) forex (i, n, 0)
 
bool cmin(ll &a, ll b) { if (b < a) { a=b; return 1; } return 0; }
bool cmax(ll &a, ll b) { if (b > a) { a=b; return 1; } return 0; }
 
const ll INF = 1e18;
const int N = 2505, LOG = 12;

struct SparseTable{
    
    vector<vector<int>> sp;
    int n;

    SparseTable (int n): n(n) { sp = vector(n+1, vector<int>(LOG, -1e9)); }
    SparseTable (int n, vector<int> &a): n(n) {
        sp = vector(n+1, vector<int>(LOG, -1e9));
        fo (i, n) sp[i][0] = a[i];
    }
    
    void build ( ) {
        fore (b, 1, LOG) {
            fo (i, n-(1<<b)+1) {
                sp[i][b] = max (sp[i][b-1], sp[i + (1<<(b-1))][b-1]);
            }
        }
    }
    
    int query (int l, int r) {
        int lg = log2(r - l + 1);
        return max (sp[l][lg], sp[r - (1<<lg)+1][lg]);
    }

};

int qans[N*N*3], L[N][N], R[N][N], pr[N][N], prc[N][N], down[N][N], up[N][N], ft[N];
vector<array<int, 4>> off[N], allt;
vector<array<int, 3>> updates, qrys;
int idx;

void update (int i, int v) {
	i++;
	for (; i < N; i += lsb(i)) 
		ft[i] += v;
}

int query (int i) {
	i++;
	int r = 0;
	for (; i > 0; i -= lsb(i))
		r += ft[i];
	return r;
}

int query (int l, int r) { if (l>r) return 0ll; return query(r) - query(l-1); }

ll count_rectangles(vector<vector<int> > a) {
	int n = a.size(), m = a[0].size();
	ll ans = 0;
	fo (j, m) {
		stack<ll> stk;
		stk.push(-1);
		fo (i, n) {
			while (stk.top() != -1 && a[stk.top()][j] < a[i][j]) stk.pop();
			up[i][j] = stk.top();
			stk.push(i);
		}
		while (stk.size()) stk.pop();
		stk.push(n);
		ffo (i, n) {
			while (stk.top() != n && a[stk.top()][j] < a[i][j]) stk.pop();
			down[i][j] = stk.top() * -1;
			stk.push(i);
		}
	}
	forex (i, n-1, 1) {
		stack<ll> stk;
		stk.push(-1);
		fo (j, m) {
			while (stk.top() != -1 && a[i][stk.top()] <= a[i][j]) stk.pop();
			L[i][j] = stk.top() + 1;
			stk.push(j);
		}
		while (stk.size()) stk.pop();
		stk.push(m);
		ffo (j, m) {
			while (stk.top() != m && a[i][stk.top()] < a[i][j]) stk.pop();
			R[i][j] = stk.top() - 1;
			stk.push(j);
		}
		fo (j, m) {
			if (L[i][j] < 1 || m-2 < R[i][j] || a[i][R[i][j]+1] == a[i][j]) continue;
			if (pr[L[i][j]][R[i][j]] == 0 || pr[L[i][j]][R[i][j]] > i+1) {
				off[i+1].pb({L[i][j], R[i][j], 1, idx});
				idx++;
			}
			off[i].pb({L[i][j], R[i][j], 1, idx});
			idx++;
			off[i-1].pb({L[i][j], R[i][j], 0, idx});
			idx++;
			pr[L[i][j]][R[i][j]] = i;
		}
	}
	SparseTable mn(m), mx(m);
	fo (i, n) {
		fo (j, m) {
			mn.sp[j][0] = down[i][j];
			mx.sp[j][0] = up[i][j];
		}
		mn.build();
		mx.build();
		for (auto e: off[i]) {
			qans[e[3]] = (e[2] == 1 ? mx.query(e[0], e[1]) : mn.query(e[0], e[1])*-1);
		}
	}
	idx = 0;
	fo (i, m) {
		fo (j, m) {
			pr[i][j] = 0;
		}
	}
	forex (i, n-1, 1) {
		fo (j, m) {
			if (L[i][j] < 1 || m-2 < R[i][j] || a[i][R[i][j]+1] == a[i][j]) continue;
			if (pr[L[i][j]][R[i][j]] == 0 || pr[L[i][j]][R[i][j]] > i+1) {
				if (qans[idx] < i) {
					allt.pb(array<int, 4>{L[i][j]*m+R[i][j], qans[idx]+1, -1, int(i+1)});
					allt.pb(array<int, 4>{L[i][j]*m+R[i][j], int(i+1), +1, int(i+1)});
				}
				idx++;
			}
			if (pr[L[i][j]][R[i][j]] > i+1 || pr[L[i][j]][R[i][j]] == 0)
				prc[L[i][j]][R[i][j]] = i; 
			if (qans[idx]+1 < i) { 
				allt.pb(array<int, 4>{L[i][j]*m+R[i][j], qans[idx]+1, -1, int(i)});
				allt.pb(array<int, 4>{L[i][j]*m+R[i][j], int(i), +1, int(i)});
			}
			idx++;
			int k = min(qans[idx], prc[L[i][j]][R[i][j]]+1);
			if (k > i)
				allt.pb(array<int, 4>{L[i][j]*m+R[i][j], int(i+1), k, -1000000000});
			idx++;
			pr[L[i][j]][R[i][j]] = i;
		}
	}
	sort(all(allt));
	// return 0;
	// for (auto q: allt) {
	// 	cout << q[0] << ' ' << q[1] << ' ' << q[2] << ' ' << q[3] << '\n';
	// }
	fo (_, allt.size()) {
		ll __ = _; 
		while (__ < allt.size() && allt[_][0] == allt[__][0]) __++;
		// cout << "Range " << l << ' ' << r << '\n';
		qrys.clear();
		fore (___, _, __) {
			if (allt[___][3] == -1e9) qrys.pb({allt[___][1], allt[___][2], allt[___][3]});
			else updates.pb({allt[___][1], allt[___][2], allt[___][3]});
		}
		reverse(all(qrys));
		// cout << qrys.size() << '\n';
		for (auto [lb, rb, ______]: qrys) {
			while (updates.size() && updates.back()[0] >= lb) {
				// cout << "Make " << updates.back()[0] << ' ' << updates.back()[1] << ' ' << updates.back()[2] << '\n';
				update(updates.back()[2], updates.back()[1]);
				updates.pop_back();
			}
			// cout << "Row " << i << ' ' << lb << ' ' << rb << ' ' << ls << '\n';
			// cout << query(lb, min(rb, ls+1)) << '\n';
			ans += query(lb, rb);
		}
		while (updates.size()) {
			update(updates.back()[2], updates.back()[1]);
			updates.pop_back();
		}
		_ = __-1;
	}
	return ans;
}

• Subtask 1: 8 / 8
• Subtask 2: 7 / 7
• Subtask 3: 12 / 12
• Subtask 4: 22 / 22
• Subtask 5: 10 / 10
• Subtask 6: 13 / 13
• Subtask 7: 28 / 28