제출 #293740

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
293740		egekabas	Rectangles (IOI19_rect)	C++14	0 / 100	69 ms	10872 KiB

rect

#include "rect.h"
#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef pair<ull, ull> pull;
typedef pair<int, int> pii;
typedef pair<ld, ld> pld;
vector<bitset<700>> root;
struct tree{
	vector<bitset<700>> seg;
	void init(int n){
		seg.resize(4*n+5);
		for(auto &u : seg)
			u.flip();
	}	
	void upd(int v, int tl, int tr, int idx, bitset<700> &val){
		if(tl == idx && tr == idx){
			seg[v] = val;
			return;
		}
		if(idx <= (tl+tr)/2)
			upd(2*v, tl, (tl+tr)/2, idx, val);
		else
			upd(2*v+1, (tl+tr)/2+1, tr, idx, val);
		seg[v] = seg[2*v]&seg[2*v+1];
	}
	int get(int v, int tl, int tr, int val){
		if(tl == tr) return tl;
		if(seg[2*v][val] == 0)
			return get(2*v, tl, (tl+tr)/2, val);
		return get(2*v+1, (tl+tr)/2+1, tr, val);
	}
};
bitset<100> lef[100][100];
bitset<100> down[100][100];
bitset<100> lefr[100][100][100];
bitset<100> downr[100][100][100];


int n, m;
long long count_rectangles(vector<std::vector<int> > a) {
	n = a.size();
	m = a[0].size();
	
	for(int i = 0; i < n; ++i)
		for(int j = 0; j < m; ++j){
			int maxi = 0;
			for(int k = j+1; k < m; ++k){
				if(a[i][j] <= maxi) break;
				lef[i][j][k] = a[i][k] > maxi;
				maxi = max(maxi, a[i][k]);
			}
		}
	
	for(int i = 0; i < n; ++i)
		for(int j = 0; j < m; ++j){
			int maxi = 0;
			for(int k = i+1; k < n; ++k){
				if(a[i][j] <= maxi) break;
				down[i][j][k] = a[k][j] > maxi;
				maxi = max(maxi, a[k][j]);
			}
		}
	
	for(int i = 0; i < n; ++i)
		for(int j = 0; j < m; ++j)
			for(int k = j; k < m; ++k){
				if(j == k)
					downr[i][j][k] = down[i][j];
				else
					downr[i][j][k] = (down[i][j]&downr[i][j][k-1]);
			}
	
	for(int i = 0; i < m; ++i)
		for(int j = 0; j < n; ++j)
			for(int k = j; k < n; ++k){
				if(j == k)
					lefr[i][j][k] = lef[j][i];
				else
					lefr[i][j][k] = (lef[j][i]&lefr[i][j][k-1]);
			}
	ll ans = 0;
	for(int i1 = 1; i1 < n-1; ++i1)
		for(int i2 = i1; i2 < n-1; ++i2)
			for(int j1 = 1; j1 < m-1; ++j1)
				for(int j2 = j1; j2 < m-1; ++j2){
					ans += downr[i1-1][j1][j2][i2+1]&lefr[j1-1][i1][i2][j2+1];
				}
	return ans;
}

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