제출 #157378

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
157378		qkxwsm	Rectangles (IOI19_rect)	C++14	100 / 100	4487 ms	784980 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<pair<short, short> > ev[MAXN][MAXN], eh[MAXN][MAXN];
int fen[MAXN];
ll ans;

void update(int idx, int v)
{
	for (int e = idx + 1; e <= max(N, M); e += e & (-e)) fen[e] += v;
}
int query(int idx)
{
	int res = 0;
	for (int e = idx + 1; e; e -= e & (-e)) res += fen[e];
	return res;
}
bool cmp(pair<short, short> a, pair<short, short> b)
{
	return (a.se < b.se);
}

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, i});
			}
			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, i});
			}
			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, i});
			}
			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, i});
			}
			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());
		}
	}
	FOR(i, 0, N)
	{
		FORD(j, M, 0)
		{
			int idx = 0;
			FOR(k, 0, SZ(ev[i][j]))
			{
				int x = ev[i][j][k].fi;
				while(idx < SZ(ev[i][j + 1]) && ev[i][j + 1][idx].fi < x) idx++;
				if (idx >= SZ(ev[i][j + 1]) || ev[i][j + 1][idx].fi != x) continue;
				ev[i][j][k].se = ev[i][j + 1][idx].se;
			}
		}
	}
	FOR(i, 0, M)
	{
		FORD(j, N, 0)
		{
			int idx = 0;
			FOR(k, 0, SZ(eh[j][i]))
			{
				int y = eh[j][i][k].fi;
				while(idx < SZ(eh[j + 1][i]) && eh[j + 1][i][idx].fi < y) idx++;
				if (idx >= SZ(eh[j + 1][i]) || eh[j + 1][i][idx].fi != y) continue;
				eh[j][i][k].se = eh[j + 1][i][idx].se;
			}
		}
	}
	//log factor here too.
	FOR(i, 1, N - 1)
	{
		FOR(j, 1, M - 1)
		{
			sort(ALL(eh[i][j]), cmp);
			for (auto p : eh[i][j])
			{
				update(p.fi, 1);
			}
			int iter = 0;
			FOR(k, 0, SZ(ev[i][j]))
			{
				while(iter < SZ(eh[i][j]) && eh[i][j][iter].se < ev[i][j][k].fi)
				{
					update(eh[i][j][iter].fi, -1);
					iter++;
				}
				ans += query(ev[i][j][k].se);
			}
			while(iter < SZ(eh[i][j]))
			{
				update(eh[i][j][iter].fi, -1);
				iter++;
			}
		}
	}
	return ans;
}

컴파일 시 표준 에러 (stderr) 메시지

rect.cpp: In function 'll count_rectangles(std::vector<std::vector<int> >)':
rect.cpp:88:12: warning: suggest parentheses around comparison in operand of '&' [-Wparentheses]
    if (idx >= 0 & idx != j - 1)
        ~~~~^~~~

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