제출 #199908

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
199908		medk	Rectangles (IOI19_rect)	C++14	72 / 100	5013 ms	475984 KiB

rect

#pragma GCC optimize ("O3")
#pragma GCC optimize ("Ofast")
#pragma GCC optimize ("unroll-loops")
 
#include <bits/stdc++.h>
#include "rect.h"
 
#define ll long long
#define pb push_back
#define x first
#define y second
 
using namespace std;
 
int where[2500][2500],where2[2500][2500];
vector<vector<pair<int,int>>> mp,mp2;
vector<pair<int,pair<int,int>>> dsu;
vector<int> sz;
 
int getp(int x)
{
    if (x==dsu[x].x)
        return x;
    return dsu[x].x=getp(dsu[x].x);
}
 
int connect(int i, int j)
{
    j=getp(j);
    if(sz[i]<sz[j]) swap(i,j);
    dsu[j].x=i;
    dsu[i].y={max(dsu[i].y.x,dsu[j].y.x),min(dsu[i].y.y,dsu[j].y.y)};
    sz[i]+=sz[j];
    return i;
}
 
ll count_rectangles(vector<vector<int>> a)
{
    for(int i=0;i<2500;i++) dsu.pb({i,{i,i}}), sz.pb(1);
    for(int i=0;i<2500;i++) for(int j=0;j<2500;j++) where[i][j]=-1,where2[i][j]=-1;
    vector<pair<int,pair<int,int>>> dsuin=dsu;
    int n=a.size(), m=a[0].size();
    vector<int> tmp(2500,0);
    for(int i=1;i<n-1;++i)
    {
        dsu=dsuin;
        fill(sz.begin(),sz.end(),1);
        vector<pair<int,int>> vp;
        for(int j=0;j<m;j++) vp.pb({a[i][j],j});
        sort(vp.begin(),vp.end());
        int startedwith=mp.size();
        for(int j=0;j<m;++j)
        {
            tmp[vp[j].y]++;
            int par=getp(vp[j].y);
            if(vp[j].y!=m-1)
                if(tmp[vp[j].y+1]==i)
                    par=connect(par, vp[j].y+1);
            if(vp[j].y!=0)
                if(tmp[vp[j].y-1]==i)
                    par=connect(par, vp[j].y-1);
            int l=dsu[par].y.x, l2=dsu[par].y.y;
            if(l2==0 || l==m-1) continue;
            if(a[i][l+1]==vp[j].x || a[i][l2-1]==vp[j].x) continue;
            int num=1;
            if(where[i-1][l+1]!=-1)
            {
                auto it = lower_bound(mp[where[i-1][l+1]].begin(),mp[where[i-1][l+1]].end(),pair<int,int>{l2-1,0});
                if(it!=mp[where[i-1][l+1]].end()) if((*it).x==l2-1) num+=(*it).y;
            }
            if(where[i][l+1]==-1) where[i][l+1]=mp.size(), mp.pb(vector<pair<int,int>>());
            mp[where[i][l+1]].pb({l2-1,num});
        }
        for(int z=startedwith;z<(int)mp.size();z++) sort(mp[z].begin(),mp[z].end());
    }
    ll ans=0;
    for(int i=0;i<2500;i++) tmp[i]=0;
    for(int i=1;i<m-1;++i)
    {
        dsu=dsuin;
        fill(sz.begin(),sz.end(),1);
        vector<pair<int,int>> vp;
        for(int j=0;j<n;++j) vp.pb({a[j][i],j});
        sort(vp.begin(),vp.end());
        int startedwith=mp2.size();
        for(int j=0;j<n;++j)
        {
            tmp[vp[j].y]++;
            int par=getp(vp[j].y);
            if(vp[j].y!=n-1)
                if(tmp[vp[j].y+1]==i)
                    par=connect(par, vp[j].y+1);
            if(vp[j].y!=0)
                if(tmp[vp[j].y-1]==i)
                    par=connect(par, vp[j].y-1);
            int l=dsu[par].y.x, l2=dsu[par].y.y;
            if(l2==0 || l==n-1) continue;
            if(a[l+1][i]==vp[j].x || a[l2-1][i]==vp[j].x) continue;
            int num=1;
            if(where2[l+1][i-1]!=-1)
            {
                auto it = lower_bound(mp2[where2[l+1][i-1]].begin(),mp2[where2[l+1][i-1]].end(),pair<int,int>{l2-1,0});
                if(it!=mp2[where2[l+1][i-1]].end()) if((*it).x==l2-1) num+=(*it).y;
            }
            if(where2[l+1][i]==-1) where2[l+1][i]=mp2.size(), mp2.pb(vector<pair<int,int>>());
            mp2[where2[l+1][i]].pb({l2-1,num});
            if(where[l][i+1]!=-1)
            {
                auto it = lower_bound(mp[where[l][i+1]].begin(),mp[where[l][i+1]].end(),pair<int,int>{i-num,l-l2+1});
                for(;it!=mp[where[l][i+1]].end();++it)
                {
                    if((*it).x>i+1) break;
                    if((*it).y>=l-l2+1) ans++;
                }
            }
        }
        for(int z=startedwith;z<(int)mp2.size();++z) sort(mp2[z].begin(),mp2[z].end());
    }
    return ans;
}

• 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: 0 / 28