제출 #143719

#제출 시각아이디문제언어결과실행 시간메모리
143719qkxwsmRectangles (IOI19_rect)C++14
72 / 100
5104 ms172424 KiB
#include "rect.h"
#include <bits/stdc++.h>

using namespace std;

template<class T, class U>
void ckmin(T &a, U b)
{
	if (a > b) a = b;
}

template<class T, class U>
void ckmax(T &a, U b)
{
	if (a < b) a = b;
}

#define y1 qkx
#define y2 wsm
#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
#define SZ(x) ((int) (x).size())
#define ALL(x) (x).begin(), (x).end()
#define INF 1000000007
#define MAXN 2513

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;

int N, M;
vector<vi> grid;
int rt[MAXN][MAXN], dn[MAXN][MAXN], lt[MAXN][MAXN], up[MAXN][MAXN];
int pref[MAXN][MAXN];
bool subtask6 = true;
vector<vector<vi> > col, row;
set<pii> edges;
ll ans;

int psum(int x1, int x2, int y1, int y2)
{
	return pref[x2 + 1][y2 + 1] - pref[x1][y2 + 1] - pref[x2 + 1][y1] + pref[x1][y1];
}
ll count_rectangles(vector<vi> a)
{
	N = SZ(a); M = SZ(a[0]); grid = a;
	FOR(i, 0, N) FOR(j, 0, M) if (grid[i][j] > 1) subtask6 = false;
	if (subtask6)
	{
		FOR(i, 0, N)
		{
			FOR(j, 0, M)
			{
				pref[i + 1][j + 1] = pref[i][j + 1] + pref[i + 1][j] - pref[i][j] + grid[i][j];
			}
		}
		//for each square, compute the next 1 on the right
		FOR(i, 0, N)
		{
			rt[i][M] = M;
			FORD(j, M, 0)
			{
				rt[i][j] = (grid[i][j] ? j : rt[i][j + 1]);
			}
		}
		FOR(i, 0, M)
		{
			dn[N][i] = N;
			FORD(j, N, 0)
			{
				dn[j][i] = (grid[j][i] ? j : dn[j + 1][i]);
			}
		}
		FOR(x1, 1, N)
		{
			FOR(y1, 1, M)
			{
				if (grid[x1][y1]) continue;
				int y2 = rt[x1][y1] - 1, x2 = dn[x1][y1] - 1;
				if (x2 == N - 1 || y2 == M - 1) continue;
				// cerr << "CONSIDER " << x1 << ' ' << y1 << " -> " << x2 << ' ' << y2 << endl;
				if (psum(x1, x2, y1, y2) != 0) continue;
				if (psum(x1 - 1, x1 - 1, y1, y2) != (y2 - y1 + 1)) continue;
				if (psum(x2 + 1, x2 + 1, y1, y2) != (y2 - y1 + 1)) continue;
				if (psum(x1, x2, y1 - 1, y1 - 1) != (x2 - x1 + 1)) continue;
				if (psum(x1, x2, y2 + 1, y2 + 1) != (x2 - x1 + 1)) continue;
				ans++;
			}
		}
	}
	else
	{
		//BE SUPER CAREFUL ABOUT EQUALITY CASE.
		//for each guy, find the next guy right of it that's greater th an it.
		//ok fix the starting column #.
		FOR(i, 0, N)
		{
			vpi cand; cand.PB({INF, M});
			FORD(j, M, 0)
			{
				while(cand.back().fi < grid[i][j]) cand.pop_back();
				rt[i][j] = cand.back().se;
				cand.PB({grid[i][j], j});
			}
			cand.clear(); cand.PB({INF, -1});
			FOR(j, 0, M)
			{
				while(cand.back().fi < grid[i][j]) cand.pop_back();
				lt[i][j] = cand.back().se;
				cand.PB({grid[i][j], j});
			}
		}
		FOR(i, 0, M)
		{
			vpi cand; cand.PB({INF, N});
			FORD(j, N, 0)
			{
				while(cand.back().fi < grid[j][i]) cand.pop_back();
				dn[j][i] = cand.back().se;
				cand.PB({grid[j][i], j});
			}
			cand.clear(); cand.PB({INF, -1});
			FOR(j, 0, N)
			{
				while(cand.back().fi < grid[j][i]) cand.pop_back();
				up[j][i] = cand.back().se;
				cand.PB({grid[j][i], j});
			}
		}
		// FOR(i, 0, N)
		// {
		// 	FOR(j, 0, M)
		// 	{
		// 		cerr << up[i][j] << ' ';
		// 	}
		// 	cerr << endl;
		// }
		FOR(i, 0, N - 1)
		{
			edges.clear();
			FOR(j, 0, M)
			{
				if (lt[i + 1][j] != -1 && lt[i + 1][j] < j - 1) edges.insert({lt[i + 1][j], j});
				if (rt[i + 1][j] != M && rt[i + 1][j] > j + 1) edges.insert({j, rt[i + 1][j]});
			}
			// cerr << "FROM ROW " << i << endl;
			FOR(j, i + 2, N)
			{
				//find all edges which they share in common.
				// cerr << "ROWS " << i << " -> " << j << endl;
				// cerr << "EDGES\n";
				// for (pii p : edges)
				// {
				// 	cerr << p.fi << ' ' << p.se << endl;
				// }
				vpi ks;
				FOR(k, 0, M)
				{
					if (dn[i][k] == j || up[j][k] == i)
					{
						// cerr << "HI\n";
						if (ks.empty() || ks.back().se != k - 1)
						{
							ks.PB({k, k});
						}
						else
						{
							ks.back().se++;
						}
						// cerr << "BYE\n";
					}
				}
				// cerr << "KS\n";
				// for (pii p : ks)
				// {
				// 	cerr << p.fi << ',' << p.se << endl;
				// }
				for (pii p : ks)
				{
					for (pii q : edges)
					{
						if (p.fi <= q.fi + 1 && q.se - 1 <= p.se)
						{
							// cerr << "YAY " << i << ' ' << j << ' ' << q.fi << ' ' << q.se << endl;
							ans++;
						}
					}
				}
				for (auto it = edges.begin(); it != edges.end(); )
				{
					int x = it -> fi, y = it -> se;
					if (rt[j][x] != y && lt[j][y] != x) it = edges.erase(it);
					else it++;
				}
				//ok now find these ranges.
			}
		}
	}
	return ans;
}
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...