제출 #145170

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
145170		jwvg0425	Rectangles (IOI19_rect)	C++17	72 / 100	5101 ms	141620 KiB

rect

#include "rect.h"
#include <stdio.h>
#include <vector>
#include <queue>
#include <algorithm>
#include <iostream>
#include <string>
#include <bitset>
#include <map>
#include <set>
#include <tuple>
#include <string.h>
#include <math.h>
#include <random>
#include <functional>
#include <assert.h>
#include <math.h>
#include <stack>
#define all(x) (x).begin(), (x).end()
#define xx first
#define yy second

using namespace std;

using i64 = long long int;
using ii = pair<int, int>;
using ii64 = pair<i64, i64>;

int up[2505][2505];
int down[2505][2505];
bool visited[2505][2505];
int dx[] = { -1, 0, 1, 0 };
int dy[] = { 0, -1, 0, 1 };

i64 sub6(vector<vector<int>> a)
{
	int n = a.size();
	int m = a[0].size();
	i64 ans = 0;

	for (int x = 1; x < n - 1; x++)
	{
		for (int y = 1; y < m - 1; y++)
		{
			if (visited[x][y] || a[x][y] == 1)
				continue;

			int lx = x, ly = y, rx = x, ry = y;
			int c = 0;

			queue<ii> q;
			q.emplace(x, y);
			visited[x][y] = true;
			while (!q.empty())
			{
				auto now = q.front();
				q.pop();
				lx = min(now.xx, lx);
				ly = min(now.yy, ly);
				rx = max(now.xx, rx);
				ry = max(now.yy, ry);
				c++;

				for (int i = 0; i < 4; i++)
				{
					int nx = now.xx + dx[i];
					int ny = now.yy + dy[i];

					if (nx == 0 || nx == n - 1 || ny == 0 || ny == m - 1 || visited[nx][ny] || a[nx][ny] == 1)
						continue;

					visited[nx][ny] = true;
					q.emplace(nx, ny);
				}
			}

			bool ok = true;
			if (c != (rx - lx + 1) * (ry - ly + 1))
				ok = false;

			for (int i = ly; i <= ry; i++)
				if (a[lx - 1][i] != 1 || a[rx + 1][i] != 1)
					ok = false;

			for (int i = lx; i <= rx; i++)
				if (a[i][ly - 1] != 1 || a[i][ry + 1] != 1)
					ok = false;

			if (ok)
				ans++;
		}
	}

	return ans;
}

i64 count_rectangles(vector<vector<int>> a)
{
	int n = a.size();
	int m = a[0].size();

	i64 ans = 0;
	int amax = 0;

	for (int x = 0; x < n; x++)
	{
		for (int y = 0; y < m; y++)
		{
			amax = max(amax, a[x][y]);
			int yu = y - 1;
			int yd = y + 1;

			while (yd < m && a[x][yd] < a[x][y])
				yd++;

			while (yu >= 0 && a[x][yu] < a[x][y])
				yu--;

			down[x][y] = yd - y - 1;
			up[x][y] = y - yu - 1;
		}
	}

	if (amax == 1)
		return sub6(a);

	for (int l = 0; l < n - 2; l++)
	{
		vector<int> dnow(m, 987654321);
		vector<int> unow(m, 987654321);
		vector<int> maxv(m, 0); // 좌우에서 y별 최댓값값

		for (int r = l + 2; r < n; r++)
		{
			vector<int> newup(m + 1, 0);
			vector<int> newdown(m + 1, 0);
			vector<int> delup(m + 1, 0);
			vector<int> deldown(m + 1, 0);
			vector<int> lrup(m + 1, 0);
			vector<int> lrdown(m + 1, 0);
			vector<bool> lr(m, 0); // l - r은 가능한지
			vector<int> depth(m + 1, 0);

			for (int y = 0; y < m; y++)
			{
				dnow[y] = min(dnow[y], down[r - 1][y]);
				unow[y] = min(unow[y], up[r - 1][y]);
				maxv[y] = max(maxv[y], a[r - 1][y]);
				lr[y] = maxv[y] < a[l][y] && maxv[y] < a[r][y];
			}

			int lru = 0, lrd = 0;

			vector<int> dval(m + 1, 0);
			vector<int> uval(m + 1, 0);

			for (int y = 0; y < m; y++)
			{
				uval[y] = min(unow[y], lru);

				if (lr[y])
					lru++;
				else
					lru = 0;
			}

			for (int y = m - 1; y >= 0; y--)
			{
				dval[y] = min(dnow[y], lrd);

				if (lr[y])
					lrd++;
				else
					lrd = 0;
			}

			for (int y = 0; y < m; y++)
			{
				if (uval[y] >= 1)
				{
					newup[y - uval[y]]++;
					delup[y - 1]++;
				}

				if (dval[y] >= 1)
				{
					newdown[y + 1]++;
					deldown[y + dval[y]]++;
				}
			}

			int d = 0;
			for (int y = 0; y < m; y++)
			{
				d += newup[y];
				d -= delup[y];

				depth[y] = d;
			}

			// (ypos, minvalue)
			stack<ii> s;

			for (int y = 1; y < m; y++)
			{
				for (int j = 0; j < newdown[y]; j++)
					s.emplace(y, 987654321);

				if (!s.empty())
				{
					s.top().yy = min(s.top().yy, depth[y]);
				}

				for (int j = 0; j < deldown[y]; j++)
				{
					ii top = s.top();
					int f = depth[top.xx - 1] + newup[top.xx];

					ans += f - top.yy;

					s.pop();
					if (!s.empty())
						s.top().yy = min(s.top().yy, top.yy);
				}
			}
		}
	}

	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