제출 #1190001

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1190001		alexdd	Rectangles (IOI19_rect)	C++20	0 / 100	76 ms	147784 KiB

rect

#include "rect.h"
#include<bits/stdc++.h>
using namespace std;

vector<pair<int,int>> get_bune(vector<int> v)
{
    vector<pair<int,int>> bune;
    deque<int> dq;
    for(int i=0;i<v.size();i++)
    {
        while(!dq.empty() && v[i] >= v[dq.back()])
        {
            if(dq.back()+1 <= i-1)
                bune.push_back({dq.back()+1,i-1});
            dq.pop_back();
        }
        if(!dq.empty() && dq.back()+1 <= i-1)
            bune.push_back({dq.back()+1,i-1});
        dq.push_back(i);
    }
    sort(bune.begin(),bune.end());
    return bune;
}
vector<pair<int,int>> bune_pe_rand[2505];
vector<int> coloane_bune[2505][2505];

long long count_rectangles(std::vector<std::vector<int>> a)
{
    long long rez=0;
    for(int i=1;i+1<a.size();i++)
    {
        bune_pe_rand[i] = get_bune(a[i]);
    }
    for(int i=1;i+1<a[0].size();i++)
    {
        vector<int> aux;
        for(int j=0;j<a.size();j++)
            aux.push_back(a[j][i]);
        vector<pair<int,int>> bune = get_bune(aux);
        for(auto [x,y]:bune)
            coloane_bune[x][y].push_back(i);
    }
    for(int sus=1;sus+1<a.size();sus++)
    {
        vector<pair<int,int>> candidati = bune_pe_rand[sus];
        for(int jos=sus;jos+1<a.size();jos++)
        {
            vector<pair<int,int>> newc;
            int pozc=0,pozr=0;
            while(pozc < candidati.size() && pozr < bune_pe_rand[jos].size())
            {
                if(candidati[pozc] == bune_pe_rand[jos][pozr])
                {
                    newc.push_back(candidati[pozc]);
                    pozc++;
                    pozr++;
                }
                else if(candidati[pozc] < bune_pe_rand[jos][pozr])
                {
                    pozc++;
                }
                else
                {
                    pozr++;
                }
            }
            candidati = newc;
            for(auto [x,y]:candidati)
            {
                int cnt=0;
                for(int c:coloane_bune[sus][jos])
                    if(x<=c && c<=y)
                        cnt++;
                if(cnt == y-x+1)
                    rez++;
            }
        }
    }
    return rez;
}
/*

ne fixam linia de sus
apoi mergem cu linia de jos in jos, o sa avem la inceput O(N) perechi de coloane candidate
cand adaugam o linie in jos o sa dezactivam niste perechi de coloane

trebuie sa verificam pt fiecare (sus,jos) care coloane sunt bune

asta o sa fie O(N^2), pt ca fiecare coloana o sa apara in O(N) (sus,jos)-uri

facem un dsu in care tinem si candidatii de perechi de coloane si cam asta e tot sper

*/

• 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