제출 #583450

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
583450		Lucpp	Rectangles (IOI19_rect)	C++17	59 / 100	3417 ms	1048576 KiB

rect

#include "rect.h"
#include <bits/stdc++.h>
using namespace std;
#define sz(x) ((int)(x).size())

vector<int> fen;
void add(int i, int x){
	for(i++; i < sz(fen); i+=i&-i) fen[i] += x;
}
int qry(int i){
	int ans = 0;
	for(i++; i > 0; i-=i&-i) ans += fen[i];
	return ans;
}
int qry(int i, int j){
	if(i > 0) return qry(j)-qry(i-1);
	else return qry(j);
}

long long count_rectangles(vector<vector<int>> a) {
	int n = sz(a), m = sz(a[0]);
	vector<vector<int>> r1(n, vector<int>(m, -1)), r2(n, vector<int>(m, -1));
	for(int i = 1; i < n-1; i++){
		stack<pair<int, int>> st({{a[i][0], 0}});
		for(int j = 1; j < m; j++){
			while(!st.empty() && a[i][j] > st.top().first){
				auto [h, k] = st.top();
				st.pop();
				r2[i][k] = j-1;
			}
			if(!st.empty()) {
				if(st.top().first > a[i][j]) r1[i][j] = st.top().second+1;
				else r1[i][j] = r1[i][st.top().second];
			}
			st.emplace(a[i][j], j);
		}
	}
	vector<vector<int>> c1(n, vector<int>(m, -1)), c2(n, vector<int>(m, -1));
	for(int j = 1; j < m-1; j++){
		stack<pair<int, int>> st({{a[0][j], 0}});
		for(int i = 1; i < n; i++){
			while(!st.empty() && a[i][j] > st.top().first){
				auto [h, k] = st.top(); st.pop();
				c2[k][j] = i-1;
			}
			if(!st.empty()){
				if(st.top().first > a[i][j]) c1[i][j] = st.top().second+1;
				else c1[i][j] = c1[st.top().second][j];
			}
			st.emplace(a[i][j], i);
		}
	}
	int mx = max(n, m);
	fen.resize(mx+1);
	vector<bool> valid(mx*mx);
	vector<vector<tuple<int, int, int>>> byI(mx*mx), byJ(mx*mx);
	set<tuple<int, int, int, int>> noDupl;
	vector<set<int>> goodI(mx*mx), goodJ(mx*mx);
	int q = 0;
	for(int i = 1; i < n-1; i++){
		for(int j = 1; j < m-1; j++){
			// cerr << "\n" << i << " " << j << " " << r1[i][j] << " " << r2[i][j] << " " << c1[i][j] << " " << c2[i][j] << "\n";
			if(c1[i][j] != -1 && c2[i][j] != -1) goodJ[c1[i][j]*mx + c2[i][j]].insert(j);
			if(r1[i][j] != -1 && r2[i][j] != -1) goodI[r1[i][j]*mx + r2[i][j]].insert(i);
			if(r1[i][j] != -1 && r2[i][j] != -1 && c1[i][j] != -1 && c2[i][j] != -1){
				auto t = make_tuple(r1[i][j], r2[i][j], c1[i][j], c2[i][j]);
				if(noDupl.count(t)) continue;
				noDupl.insert(t);
				valid[q] = true;
				byI[c1[i][j]*mx + c2[i][j]].emplace_back(r1[i][j], r2[i][j], q);
				byJ[r1[i][j]*mx + r2[i][j]].emplace_back(c1[i][j], c2[i][j], q);
				q++;
			}
		}
	}
	for(int L = 1; L < n-1; L++){
		for(int R = L; R < n-1; R++){
			// cerr << "# " << L << " " << R << "\n";
			for(int j : goodJ[L*mx+R]){
				// cerr << j << "\n";
				add(j, 1);
			}
			for(auto [a, b, ind] : byI[L*mx+R]){
				// cerr << a << " " << b << " " << qry(a, b) << "\n";
				if(qry(a, b) != b-a+1) valid[ind] = false;
			}
			for(int j : goodJ[L*mx+R]){
				add(j, -1);
			}
		}
	}
	for(int L = 1; L < m-1; L++){
		for(int R = L; R < m-1; R++){
			for(int i : goodI[L*mx+R]){
				add(i, 1);
			}
			for(auto [a, b, ind] : byJ[L*mx+R]){
				if(qry(a, b) != b-a+1) valid[ind] = false;
			}
			for(int i : goodI[L*mx+R]){
				add(i, -1);
			}
		}
	}
	int ans = 0;
	for(bool b : valid) ans += b;
	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