제출 #651932

#제출 시각아이디문제언어결과실행 시간메모리
651932jiahngArranging Shoes (IOI19_shoes)C++14
100 / 100
135 ms141304 KiB
#include "shoes.h"
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int,int> pi;
typedef vector <int> vi;
typedef vector <pi> vpi;
typedef pair<pi, ll> pii;
typedef set <ll> si;
typedef long double ld;
#define f first
#define s second
#define mp make_pair
#define FOR(i,s,e) for(int i=s;i<=int(e);++i)
#define DEC(i,s,e) for(int i=s;i>=int(e);--i)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define lbd(x, y) lower_bound(all(x), y)
#define ubd(x, y) upper_bound(all(x), y)
#define aFOR(i,x) for (auto i: x)
#define mem(x,i) memset(x,i,sizeof x)
#define fast ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define maxn 200010
#define INF (ll)1e9
#define MOD 1000000007
typedef pair <vi, int> pvi;
typedef pair <int,pi> ipi;
typedef vector <pii> vpii;

int N;
queue <int> A[maxn];
int fw[maxn];
int qry(int p){
    int ans = 0;
    for (int i = p; i > 0; i -= i & (-i)) ans += fw[i];
    return ans;
}
void upd(int p, int v){
    for (int i = p; i <= N; i += i & (-i)) fw[i] += v;
}
long long count_swaps(std::vector<int> s) {
    vpi v;
    N = s.size();
    FOR(i,1,s.size()){
       int val = (s[i-1] < 0 ? -i : i);
       //cout << val << ' ';
       if (A[abs(s[i-1])].empty() || (A[abs(s[i-1])].front() < 0) == (val < 0)){
            A[abs(s[i-1])].push(val); 
       }else{
            if (val < 0){
                v.pb(pi(-val, A[abs(s[i-1])].front()));
            }else{
                
                v.pb(pi(-A[abs(s[i-1])].front(),val));
            }
            A[abs(s[i-1])].pop();
        }
    }
    ll ans = 0;
    FOR(i,0,v.size()-1){
        if (v[i].f > v[i].s){
            ans++; swap(v[i].f,v[i].s);    
        }
    }
    sort(all(v));
    /*
    FOR(i,0,v.size()-1) FOR(j,i+1,v.size()-1){
        if (v[j].s < v[i].s) ans += 2;
        else if (v[j].f < v[i].s) ans++;
    }
    return ans;
*/
    /*
    sort(all(v), [](pi i,pi j){
            
        return (i.f < j.f) + (i.f < j.s) + (i.s < j.f) + (i.s < j.s) >= 2;
    });
    */

    vi v2;
    aFOR(i, v){
        v2.pb(i.f); v2.pb(i.s);
    }
    aFOR(i, v2){
        ans += qry(N) - qry(i);
        upd(i, 1);
    }
	return ans;
}
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