제출 #1032350

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1032350		Kipras	Arranging Shoes (IOI19_shoes)	C++17	100 / 100	204 ms	153652 KiB

shoes

#include <bits/stdc++.h>
 
typedef long long ll;
 
using namespace std;
 
const ll maxN = 1e5+10;
 
ll n;
 
ll val[maxN*4], l[maxN*4], r[maxN*4], lazy[maxN*4];
deque<ll> pos[maxN], neg[maxN];
ll used[maxN*2]={0};
ll ind = 0;
 
void build(ll v){
 
    if(v>=n){
        l[v]=ind;
        r[v]=ind;
        ind++;
        val[v]=0;
        lazy[v]=0;
    }else{
        build(v*2);
        build(v*2+1);
        l[v]=l[v*2];
        r[v]=r[v*2+1];
        val[v]=0;
        lazy[v]=0;
    }
 
}
 
void update(ll v, ll L, ll R, ll x){
 
    if(L>r[v]||R<l[v]){
        //cout<<"ending: "<<l[v]<<" "<<r[v]<<" "<<L<<" "<<R<<" "<<v<<"\n";
        return;
    }
    else if(L<=l[v]&&r[v]<=R){
        lazy[v]+=x;
        //cout<<"updation: "<<l[v]<<" "<<r[v]<<" "<<v<<"\n";
    }else{
        update(v*2, L, R, x);
        update(v*2+1, L, R, x);
    }
 
}
 
ll query(ll v, ll L, ll R){
 
    if(L>r[v]||R<l[v]){
        return 0;
    }
    else if(L<=l[v]&&r[v]<=R){
        val[v]+=lazy[v];
        lazy[v]=0;
        return val[v];
    }else{
        lazy[v*2]+=lazy[v];
        lazy[v*2+1]+=lazy[v];
        lazy[v]=0;
        return query(v*2, L, R)+query(v*2+1, L, R);
    }
 
}
 
ll count_swaps(vector<int> s){
 
    n = s.size();
 
    build(1);
 
    ll rez = 0;
 
    for(int i = 0; i < n; i++){
        if(s[i]>0)
            pos[s[i]].push_back(i);
        else
            neg[abs(s[i])].push_back(i);
    }
 
    for(int i = 0; i < n; i++){
        if(used[i])continue;
 
        ll v=abs(s[i]);
        if(s[i]>0){
            rez++;
            pos[v].pop_front();
            ll u = neg[v].front();
            neg[v].pop_front();
            ll p1 = i + query(1, i, i);
            ll p2 = u + query(1, u, u);
            update(1, i, u, 1);
            //cout<<rez<<" "<<i<<" "<<u<<" "<<p1<<" "<<p2<<" a\n";
            rez+=p2-p1-1;
            //cout<<rez<<"a\n";
            used[u]=1;
        }else{
            neg[v].pop_front();
            ll u = pos[v].front();
            pos[v].pop_front();
            ll p1 = i + query(1, i, i);
            ll p2 = u + query(1, u, u);
            update(1, i, u, 1);
            //cout<<rez<<" "<<i<<" "<<u<<" "<<p1<<" "<<p2<<" b\n";
            rez+=p2-p1-1;
            //cout<<rez<<"b\n";
            used[u]=1;
        }
 
    }
 
    return rez;
 
}
 

• Subtask 1: 10 / 10
• Subtask 2: 20 / 20
• Subtask 3: 20 / 20
• Subtask 4: 15 / 15
• Subtask 5: 20 / 20
• Subtask 6: 15 / 15