제출 #276192

#제출 시각아이디문제언어결과실행 시간메모리
276192Toirov_SadiArranging Shoes (IOI19_shoes)C++17
25 / 100
230 ms17612 KiB
#include<bits/stdc++.h>
#include "shoes.h"

using namespace std;

const int N = 2e5 + 7;

int t[4 * N];
int lz[4 * N];
void push(int v, int tl, int tr){
    if(lz[v] == 0) return;
    int tm = (tl + tr) / 2;
    t[v * 2] += (tm - tl + 1) * lz[v];
    t[v * 2 + 1] += (tr - tm) * lz[v];
    lz[v * 2] += lz[v];
    lz[v * 2 + 1] += lz[v];
    lz[v] = 0;
}
void upd(int v, int tl, int tr, int l, int r, int x){
    if(l > r) return;

    if(tl == l && tr == r){
        t[v] += (tr - tl + 1) * x;
        lz[v] += x;
        return;
    }
    push(v, tl, tr);
    int tm = (tl + tr) / 2;
    upd(v * 2, tl, tm, l, min(tm, r), x);
    upd(v * 2 + 1, tm + 1, tr, max(tm + 1, l), r, x);
    t[v] = t[v * 2] + t[v * 2 + 1];
}
int get(int v, int tl, int tr, int l, int r){
    if(l > r) return 0;
    if(tl == l && tr == r){
        return t[v];
    }
    push(v, tl, tr);
    int tm = (tl + tr) / 2;
    return (get(v * 2, tl, tm, l, min(tm, r)) +
            get(v * 2 + 1, tm + 1, tr, max(tm + 1, l), r));
}

long long count_swaps(vector<int> s) {
    int n = (int)s.size();

    priority_queue<int> g[(n / 2) + 1][2];

    vector<pair<int, int>> a;
    for(int i = 0; i < (int)s.size(); i ++){
        int t = (s[i] > 0);

        int x = abs(s[i]);

        if(!g[x][1 - t].empty()){
            int y = -g[x][1 - t].top();
            g[x][1 - t].pop();
            a.push_back({y, i});
            int dist = i - y - 1 + (s[y] > s[i]);
            upd(1, 1, n, y + 1, y + 1, dist);
            upd(1, 1, n, i + 1, i + 1, dist);
        }
        else{
            g[x][t].push(-i);
        }
    }

    sort(a.rbegin(), a.rend(), [](pair<int, int> x, pair<int, int> y){
        int dist1 = x.second - x.first;
        int dist2 = y.second - y.first;
        if(dist1 == dist2){
            return (x.first < y.first);
        }
        return (dist1 < dist2);
    });

    long long res = 0;
    while(!a.empty()){
        pair<int, int> cur = a.back();
        a.pop_back();
        int l = cur.first + 1;
        int r = cur.second + 1;
        res += max(0, min(get(1, 1, n, l, l), get(1, 1, n, r, r)));

        upd(1, 1, n, l, r, -1);
    }
    return res;
}
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