제출 #670049

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
670049		illyakr	Arranging Shoes (IOI19_shoes)	C++14	100 / 100	126 ms	24060 KiB

shoes

//#pragma GCC optimize("inline,Ofast,no-stack-protector,unroll-loops,fast-math,O3")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,popcnt,avx,abm")
#include <bits/stdc++.h>
#include "shoes.h"

using namespace std;

int t[801010];
int tAdd[801010];
void build(int v, int vl, int vr) {
    if (vl == vr) {
        t[v] = vl;
        return;
    }
    int vm = (vl + vr) / 2;
    build(2 * v, vl, vm);
    build(2 * v + 1, vm + 1, vr);
}
void push(int v, int vl, int vr) {
    if (tAdd[v] == 0)return;
    t[v] += tAdd[v];
    if (vl != vr) {
        tAdd[2 * v] += tAdd[v];
        tAdd[2 * v + 1] += tAdd[v];
    }
    tAdd[v] = 0;
    return;
}
void upd(int v, int vl, int vr, int l, int r, int val) {
    push(v, vl, vr);
    if (l <= vl && vr <= r) {
        tAdd[v] += val;
        push(v, vl, vr);
        return;
    }
    if (r < vl || vr < l)return;
    int vm = (vl + vr) / 2;
    upd(2 * v, vl, vm, l, r, val);
    upd(2 * v + 1, vm + 1, vr, l, r, val);
}
int gt(int v, int vl, int vr, int pos) {
    push(v, vl, vr);
    if (vl == vr)
        return t[v];
    int vm = (vl + vr) / 2;
    if (pos <= vm)
        return gt(2 * v, vl, vm, pos);
    return gt(2 * v + 1, vm + 1, vr, pos);
}

bool srt[401010];
bool used[401010];
vector<int> have[401010];
int p[401010];
long long count_swaps(vector<int> s) {
    int n = s.size();
    build(1, 0, n - 1);
    for (int i = 0; i < n; i++)
        have[s[i] + 200000].push_back(i);
    long long ans = 0;
    for (int i = 0; i < n; i++) {
        if (used[i])continue;
        long long val = -s[i] + 200000;
        if (!srt[val]) {
            srt[val] = true;
            sort(have[val].begin(), have[val].end());
        }
        long long sc = have[val][p[val]++];
        p[s[i] + 200000]++;

        long long cur = gt(1, 0, n - 1, i);
        long long id = gt(1, 0, n - 1, sc);

//        cout << i << ", " << cur << " <<  " << sc << ", " << id << endl;
        used[sc] = true;

        ans += (id - cur);
        if (s[i] < 0)
            ans--;
        upd(1, 0, n - 1, i, sc, 1);
    }
    return ans;
}

/**
2
2 1 -1 -2

3
-2 2 2 -2 -2 2
*/

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