제출 #1068090

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1068090		Gromp15	Rectangles (IOI19_rect)	C++17	72 / 100	1626 ms	242772 KiB

rect

#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("avx2")
#include "rect.h"
#define ll long long
#define ar array
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
using namespace std;
template<typename T> bool ckmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }

using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define rint(l, r) uniform_int_distribution<int>(l, r)(rng)

struct dsu {
	int N, M;
	vector<int> p, size, lx, rx, ly, ry;
	dsu(int n, int m) : N(n), M(m), p(n*m), size(n*m, 1), lx(n*m), rx(n*m), ly(n*m), ry(n*m) {
		for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) {
			p[i*m+j] = i*m+j;
			lx[i*m+j] = i;
			rx[i*m+j] = i;
			ly[i*m+j] = j;
			ry[i*m+j] = j;
		}
	}
	int get(int v) {
		return v == p[v] ? v : p[v] = get(p[v]);
	}
	void merge(int a, int b) {
		a = get(a), b = get(b);
		if (a == b) return;
		if (size[a] < size[b]) swap(a, b);
		ckmin(lx[a], lx[b]);
		ckmin(ly[a], ly[b]);
		ckmax(rx[a], rx[b]);
		ckmax(ry[a], ry[b]);
		size[a] += size[b], p[b] = a;
	}
	int good(int v) {
		return v == get(v) && (rx[v] - lx[v] + 1) * (ry[v] - ly[v] + 1) == size[v] && lx[v] >= 1 && ly[v] >= 1 && rx[v] < N-1 && ry[v] < M-1;
	}
};

long long count_rectangles(std::vector<std::vector<int>> a) {
	int n = sz(a), m = sz(a[0]);
	if ((n <= 700 && m <= 700) || n <= 3) {
		int n = sz(a), m = sz(a[0]);
		vector ok(n, vector(n, vector<bool>(m)));
		for (int i = 0; i < m; i++) {
			for (int j = 1; j < n-1; j++) {
				int mx = 0;
				for (int k = j; k < n-1; k++) {
					ckmax(mx, a[k][i]);
					ok[j][k][i] = mx >= min(a[j-1][i], a[k+1][i]);
				}
			}
		}
		vector<vector<int>> right(n, vector<int>(m, -1)), left(n, vector<int>(m, m-1));
		for (int i = 0; i < n; i++) {
			for (int j = 1; j < m-1; j++) {
				int mx = 0;
				for (int k = j; k < m-1; k++) {
					ckmax(mx, a[i][k]);
					if (mx >= a[i][j-1]) break;
					right[i][j] = k;
				}
			}
			for (int j = m-2; j >= 1; j--) {
				int mx = 0;
				for (int k = j; k >= 1; k--) {
					ckmax(mx, a[i][k]);
					if (mx >= a[i][j+1]) break;
					left[i][j] = k;
				}
			}
		}
		long long ans = 0;
		for (int i = 1; i < n-1; i++) {
			vector<int> R(m, m-2), L(m, -1);
			for (int j = i; j < n-1; j++) {
				for (int k = 0; k < m; k++) {
					ckmin(R[k], right[j][k]);
					ckmax(L[k], left[j][k]);
				}
				vector<int> to(m, -1);
				for (int k = m-1; k >= 0; k--) if (!ok[i][j][k]) {
					int l = k;
					while (l && !ok[i][j][l-1]) l--;
					for (int z = l; z <= k; z++) to[z] = k;
					k = l;
				}
				for (int k = 1; k < m-1; k++) {
					int r = min(to[k], R[k]);
					for (int l = k; l <= r; l++) ans += L[l] <= k;
				}
			}
		}
		return ans;
	}
	dsu D(n, m);
	for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) if (!a[i][j]) {
		if (i+1 < n && !a[i+1][j]) D.merge(i*m+j, (i+1)*m+j);
		if (j+1 < m && !a[i][j+1]) D.merge(i*m+j, i*m+j+1);
	}
	int ans = 0;
	for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) ans += !a[i][j] && D.good(i*m+j);
	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: 0 / 28