Submission #1190058

#	Time	Username	Problem	Language	Result	Execution time	Memory
1190058		alexdd	Rectangles (IOI19_rect)	C++20	100 / 100	2549 ms	866676 KiB

rect

#pragma GCC optimize("O3,unroll-loops")
#include "rect.h"
#include<iostream>
#include<algorithm>
using namespace std;

vector<pair<int,int>> get_bune(vector<int> v)
{
    vector<pair<int,int>> bune;
    vector<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});
        if(!dq.empty() && v[dq.back()] == v[i])
            dq.pop_back();
        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];
vector<int> randuri_bune[2505][2505],primu[2505][2505];
int poz_rbune[2505][2505];

int aib[2505][2505],m;
int qry(int x, int y)
{
    x++;
    y++;
    int aux=0;
    for(int i=x;i<=m;i+=(i&(-i)))
        for(int j=y;j>0;j-=(j&(-j)))
            aux += aib[i][j];
    return aux;
}
void upd(int x, int y, int newv)
{
    x++;
    y++;
    for(int i=x;i>0;i-=(i&(-i)))
        for(int j=y;j<=m;j+=(j&(-j)))
            aib[i][j] += newv;
}

vector<pair<int,int>> scoase[2505];
long long count_rectangles(std::vector<std::vector<int>> a)
{
    m = a[0].size();
    long long rez=0;
    for(int i=1;i+1<a.size();i++)
    {
        bune_pe_rand[i] = get_bune(a[i]);
        for(auto [x,y]:bune_pe_rand[i])
            randuri_bune[x][y].push_back(i);
    }
    for(int x=1;x+1<m;x++)
    {
        for(int y=x;y+1<m;y++)
        {
            if(randuri_bune[x][y].empty())
                continue;
            primu[x][y].resize(randuri_bune[x][y].size() + 2);
            int ult = a.size();
            if(randuri_bune[x][y].back() != (int)a.size() - 2)
                ult = randuri_bune[x][y].back() + 1;
            primu[x][y][randuri_bune[x][y].size()] = ult;
            for(int i=randuri_bune[x][y].size()-1;i>0;i--)
            {
                if(randuri_bune[x][y][i] != randuri_bune[x][y][i-1] + 1)
                    ult = randuri_bune[x][y][i-1] + 1;
                primu[x][y][i] = ult;
            }
        }
    }
    for(int i=1;i+1<m;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++)
    {
        for(int jos=sus;jos<=a.size();jos++)
            scoase[jos].clear();
        for(auto [x,y]:bune_pe_rand[sus])
        {
            upd(x,y,+1);
            while(randuri_bune[x][y][poz_rbune[x][y]] < sus)
                poz_rbune[x][y]++;
            scoase[primu[x][y][poz_rbune[x][y]+1]].push_back({x,y});
        }
        for(int jos=sus;jos+1<a.size();jos++)
        {
            //for(int i=0;i<scoase[jos].size();i++) upd(scoase[jos][i].first,scoase[jos][i].second,-1);
            for(auto [x,y]:scoase[jos]) upd(x,y,-1);
            //for(pair<int,int> x:scoase[jos]) upd(x.first,x.second,-1);

            /*int ult=0;
            for(int i=0;i<coloane_bune[sus][jos].size();i++)
            {
                if(i == (int)coloane_bune[sus][jos].size() - 1 || coloane_bune[sus][jos][i] + 1 != coloane_bune[sus][jos][i+1])
                {
                    int le = coloane_bune[sus][jos][ult], ri = coloane_bune[sus][jos][i];
                    rez += qry(le,ri);
                    ult = i+1;
                }
            }*/
            if(!coloane_bune[sus][jos].empty())
            {
                int ult = coloane_bune[sus][jos][0];
                int prec = ult-1;
                for(int x:coloane_bune[sus][jos])
                {
                    if(x != prec+1)
                    {
                        rez += qry(ult,prec);
                        ult = x;
                    }
                    prec = x;
                }
                rez += qry(ult,prec);
            }

        }
        for(auto [x,y]:scoase[a.size()]) upd(x,y,-1);
    }
    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: 8 / 8
• Subtask 2: 7 / 7
• Subtask 3: 12 / 12
• Subtask 4: 22 / 22
• Subtask 5: 10 / 10
• Subtask 6: 13 / 13
• Subtask 7: 28 / 28