Submission #143720

#	Time	Username	Problem	Language	Result	Execution time	Memory
143720		qkxwsm	Rectangles (IOI19_rect)	C++14	59 / 100	5034 ms	171712 KiB

rect

#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;
	//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;
}

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