Submission #1067996

#	Time	Username	Problem	Language	Result	Execution time	Memory
1067996		Gromp15	Rectangles (IOI19_rect)	C++17	0 / 100	12 ms	348 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 fenwick {
	int n; vector<int> bit;
	fenwick(int a) : n(a), bit(a+1) {}
	void update(int pos, int x) {
		pos++;
		for (int i = pos; i <= n; i += i&-i) bit[i] += x;
	}
	int sum(int pos) {
		int res = 0;
		while (pos) res += bit[pos], pos -= pos&-pos;
		return res;
	}
	int query(int l, int r) {
		return sum(r+1) - sum(l);
	}
};

long long count_rectangles(std::vector<std::vector<int>> a) {
	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;
			}
			fenwick fw(m);
			for (int k = 1; k < m-1; k++) {
				int r = min(to[k], R[k]);
				if (r >= k) fw.update(r, 1), fw.update(k-1, -1);
				if (L[k] <= k) ans += fw.query(L[k], k);
			}
		}
	}
	return ans;
}

• Subtask 1: 0 / 8
• Subtask 2: 0 / 7
• Subtask 3: 0 / 12
• Subtask 4: 0 / 22
• Subtask 5: 0 / 10
• Subtask 6: 0 / 13
• Subtask 7: 0 / 28