제출 #588297

#제출 시각아이디문제언어결과실행 시간메모리
588297perchutsBubble Sort 2 (JOI18_bubblesort2)C++17
100 / 100
3687 ms129576 KiB
#include <bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define sz(x) (int) x.size()
#define endl '\n'
#define pb push_back
#define _ ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
using namespace std;
 
using ll = long long;
using ull = unsigned long long;
using ii = pair<int,int>;
using iii = tuple<int,int,int>;
 
const int inf = 2e9+1;
const int mod = 1e9+7;
const int maxn = 1e6+100;
 
template<typename X, typename Y> bool ckmin(X& x, const Y& y) { return (y < x) ? (x=y,1):0; }
template<typename X, typename Y> bool ckmax(X& x, const Y& y) { return (x < y) ? (x=y,1):0; }
 
int val[maxn], seg[4*maxn], lazy[4*maxn], n, q, freq[4*maxn];
 
map<ii,int>v2;
 
//comprimir coordenadas de todos os valores (iniciais e alteracoes nas queries)
//necessario pra ordenar: max(0,posicao atual - posicao ordenada)
//pra achar posicao ordenada: seg de frequencia com compressao
//update: vou trocar x por y
//se x < y, todo mundo entre compressao(x) e compressao(y) diminui em 1.
//se x > y, todo mundo entre compressao(x) e compressao(y) aumenta em 1.
//se x=y, fodase
 
void updatef(int i,int l,int r,int x,int d){
    if(l>x||r<x)return;
    if(l==r){
        freq[i] = d;
        return;
    }
    int md = (l+r)/2;
    updatef(2*i,l,md,x,d), updatef(2*i+1,md+1,r,x,d);
    freq[i] = freq[2*i] + freq[2*i+1];
}
 
int query(int i,int l,int r,int x){
    if(!x)return 0;
    if(l==r)return freq[i];
    int md = (l+r)/2;
    if(md<x)return freq[2*i] + query(2*i+1,md+1,r,x);
    return query(2*i,l,md,x);
}
 
void push(int i,int l,int r){
    seg[i] += lazy[i];
    if(l!=r)lazy[2*i] += lazy[i], lazy[2*i+1] += lazy[i];
    lazy[i] = 0;
}
 
void adiciona(int i,int l,int r,int x,int d){
    push(i,l,r);
    if(l>x||r<x)return;
    if(l==r){
        seg[i] = d;
        return;
    } 
    int md = (l+r)/2;
    adiciona(2*i,l,md,x,d), adiciona(2*i+1,md+1,r,x,d);
    seg[i] = max(seg[2*i],seg[2*i+1]);
}
 
void update(int i,int l,int r,int x,int y,int d){
    push(i,l,r);
    if(l>y||r<x)return;
    if(x<=l&&r<=y){
        lazy[i] += d;
        push(i,l,r);
        return;
    }
    int md = (l+r)/2;
    update(2*i,l,md,x,y,d), update(2*i+1,md+1,r,x,y,d);
    seg[i] = max(seg[2*i],seg[2*i+1]);
}
 
 
vector<int> countScans(vector<int> a,vector<int> x,vector<int> v){
	q=sz(x), n=sz(a);
 
    vector<int>answers;
 
    for(int i=1;i<=4*(n+q);++i)seg[i] = -6e5, freq[i] = 0;
 
    for(int i=0;i<n;++i){
        v2[{a[i],i}] = 0;
    }
 
    for(int i=0;i<q;++i){
        v2[{v[i],x[i]}] = 0;
    }

    int qnt = 1;

    for(auto& [k,l]:v2)l = qnt++;
 
    for(int i=0;i<n;++i){
        a[i] = v2[{a[i],i}];
    }

    for(int i=0;i<q;++i){
        v[i] = v2[{v[i],x[i]}];
    }

    for(int i=0;i<n;++i){
        updatef(1,1,qnt,a[i],1);
    }

    for(int i=0;i<n;++i){
        adiciona(1,1,qnt,a[i],max(0,i - query(1,1,qnt,a[i]-1)));
    }
 
    for(int i=0;i<q;++i){
        int atual = a[x[i]], novo = v[i];
 
        updatef(1,1,qnt,atual,0), adiciona(1,1,qnt,atual,-6e5), updatef(1,1,qnt,novo,1);
 
        update(1,1,qnt,atual+1,qnt,1);
        update(1,1,qnt,novo+1,qnt,-1);

        adiciona(1,1,qnt,novo,max(0,x[i] - query(1,1,qnt,v[i]-1)));
        push(1,1,qnt);
        answers.pb(max(0,seg[1]));
        a[x[i]] = novo;
    }
    return answers;
}
 
//5 4 3 2 5
//4 3 2 5 5
//3 2 4 5 5
 
 
// int main(){
//     int n,q;cin>>n>>q;
//     vector<int>a;
//     for(int i=1;i<=n;++i){
//         int x;cin>>x;
//         a.pb(x);
//     }
//     vector<int>x,v;
//     for(int i=0;i<q;++i){
//         int d,f;cin>>d>>f;
//         x.pb(d), v.pb(f);
//     }
//     vector<int>resp = countScans(a,x,v);
//     for(auto x:resp)cout<<x<<" ";
// }
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