제출 #157368

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
157368		qkxwsm	Rectangles (IOI19_rect)	C++14	10 / 100	577 ms	594224 KiB

rect

#include "rect.h"
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;

template<class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

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<short> ev[MAXN][MAXN], eh[MAXN][MAXN], dv[MAXN][MAXN], dh[MAXN][MAXN];
ll ans;

ll count_rectangles(vector<vi> a)
{
	N = SZ(a); M = SZ(a[0]);
	FOR(i, 0, N)
	{
		vpi cand; cand.PB({INF, M + 1});
		FORD(j, M, 0)
		{
			while(cand.back().fi < a[i][j]) cand.pop_back();
			short idx = cand.back().se;
			if (idx < M && idx != j + 1)
			{
				eh[i][j + 1].PB(idx - 1);
			}
			cand.PB({a[i][j], j});
		}
		cand.clear(); cand.PB({INF, -2});
		FOR(j, 0, M)
		{
			while(cand.back().fi < a[i][j]) cand.pop_back();
			short idx = cand.back().se;
			if (idx >= 0 & idx != j - 1)
			{
				eh[i][idx + 1].PB(j - 1);
			}
			cand.PB({a[i][j], j});
		}
	}
	FOR(i, 0, M)
	{
		vpi cand; cand.PB({INF, N + 1});
		FORD(j, N, 0)
		{
			while(cand.back().fi < a[j][i]) cand.pop_back();
			short idx = cand.back().se;
			if (idx < N && idx != j + 1)
			{
				ev[j + 1][i].PB(idx - 1);
			}
			cand.PB({a[j][i], j});
		}
		cand.clear(); cand.PB({INF, -2});
		FOR(j, 0, N)
		{
			while(cand.back().fi < a[j][i]) cand.pop_back();
			short idx = cand.back().se;
			if (idx >= 0 && idx != j - 1)
			{
				ev[idx + 1][i].PB(j - 1);
			}
			cand.PB({a[j][i], j});
		}
	}
	FOR(i, 0, N)
	{
		FOR(j, 0, M)
		{
			// sort(ALL(ev[i][j])); ev[i][j].erase(unique(ALL(ev[i][j])), ev[i][j].end());
			// sort(ALL(eh[i][j])); eh[i][j].erase(unique(ALL(eh[i][j])), eh[i][j].end());
			dv[i][j].resize(SZ(ev[i][j]));
			dh[i][j].resize(SZ(eh[i][j]));
		}
	}
	FOR(i, 0, N)
	{
		FORD(j, M, 0)
		{
			int idx = 0;
			FOR(k, 0, SZ(ev[i][j]))
			{
				dv[i][j][k] = j;
				int x = ev[i][j][k];
				while(idx < SZ(ev[i][j + 1]) && ev[i][j + 1][idx] < x) idx++;
				if (idx >= SZ(ev[i][j + 1]) || ev[i][j + 1][idx] != x) continue;
				dv[i][j][k] = dv[i][j + 1][idx];
			}
		}
	}
	FOR(i, 0, M)
	{
		FORD(j, N, 0)
		{
			int idx = 0;
			FOR(k, 0, SZ(eh[j][i]))
			{
				dh[j][i][k] = j;
				int y = eh[j][i][k];
				while(idx < SZ(eh[j + 1][i]) && eh[j + 1][i][idx] < y) idx++;
				if (idx >= SZ(eh[j + 1][i]) || eh[j + 1][i][idx] != y) continue;
				dh[j][i][k] = dh[j + 1][i][idx];
			}
		}
	}
	//log factor here too.
	FOR(i, 1, N - 1)
	{
		FOR(j, 1, M - 1)
		{
			vpi ps;
			FOR(k, 0, SZ(eh[i][j]))
			{
				ps.PB({dh[i][j][k], eh[i][j][k]});
			}
			sort(ALL(ps));
			ordered_set<pii> ord;
			FOR(k, 0, SZ(ps))
			{
				ord.insert({ps[k].se, k});
			}
			int iter = 0;
			FOR(k, 0, SZ(ev[i][j]))
			{
				while(iter < SZ(ps) && ps[iter].fi < ev[i][j][k])
				{
					ord.erase({ps[iter].se, iter});
					iter++;
				}
				ans += ord.order_of_key({dv[i][j][k], INF});
			}
			ord.clear();
		}
	}
	return ans;
}

컴파일 시 표준 에러 (stderr) 메시지

rect.cpp: In function 'll count_rectangles(std::vector<std::vector<int> >)':
rect.cpp:72:12: warning: suggest parentheses around comparison in operand of '&' [-Wparentheses]
    if (idx >= 0 & idx != j - 1)
        ~~~~^~~~

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