제출 #563229

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
563229		Valaki2	Arranging Shoes (IOI19_shoes)	C++14	45 / 100	95 ms	16044 KiB

shoes

#include "shoes.h"
#include <bits/stdc++.h>
using namespace std;

#define int long long
#define pb push_back
#define mp make_pair

int n;
vector<int> val;
vector<int> diff; // diff[i] = value(i) - value(i - 1)
vector<int> segtree;

int range_query(int v, int vl, int vr, int ql, int qr) {
    if(ql > vr || qr < vl) {
        return 0;
    }
    if(vl == ql && vr == qr) {
        return segtree[v];
    }
    int mid = (vl + vr) / 2;
    return range_query(2 * v, vl, mid, ql, min(qr, mid))
            + range_query(2 * v + 1, mid + 1, vr, max(ql, mid + 1), qr);
}

int query(int pos) {
    return range_query(1, 1, 2 * n + 1, 1, pos);
}

void point_update(int v, int vl, int vr, int pos, int add) {
    if(vl == vr) {
        segtree[v] += add;
    } else {
        int mid = (vl + vr) / 2;
        if(pos <= mid) {
            point_update(2 * v, vl, mid, pos, add);
        } else {
            point_update(2 * v + 1, mid + 1, vr, pos, add);
        }
        segtree[v] = segtree[2 * v] + segtree[2 * v + 1];
    }
}

void update(int l, int r, int add) {
    point_update(1, 1, 2 * n + 1, l, add);
    point_update(1, 1, 2 * n + 1, r + 1, -add);
}

int solve() {
    int ans = 0;
    diff.assign(1 + 2 * n + 1, 0);
    segtree.assign(1 + 4 * (2 * n + 1), 0);
    vector<int> where(1 + 2 * n, 0);
    vector<int> nxt(1 + 2 * n, 0);
    vector<int> cur_pos(1 + n, 0);
    for(int i = 2 * n; i >= 1; i--) {
        if(val[i] > 0) {
            nxt[i] = cur_pos[val[i]];
            cur_pos[val[i]] = i;
        }
    }
    int cnt = 0;
    for(int i = 1; i <= 2 * n; i++) {
        if(val[i] < 0) {
            //cout << i << ": " << ans << " ";
            cnt++;
            ans += (i + query(i)) - (2 * cnt - 1);
            //cout << ans << " ";
            // upd between 2 * cnt - 1 and i
            update(1, i, 1);
            ans += (cur_pos[-val[i]] + query(cur_pos[-val[i]])) - (2 * cnt);
            //cout << ans << "\n";
            // upd between 2 * cnt and cur_pos[-val[i]]
            update(1, cur_pos[-val[i]], 1);
            cur_pos[-val[i]] = nxt[cur_pos[-val[i]]];
        }
    }
    return ans;
}

long long count_swaps(vector<signed> s) {
    n = s.size() / 2;
    val.pb(0);
    for(int i = 0; i < 2 * n; i++) {
        val.pb(s[i]);
    }
	return solve();
}

• Subtask 1: 10 / 10
• Subtask 2: 0 / 20
• Subtask 3: 20 / 20
• Subtask 4: 15 / 15
• Subtask 5: 0 / 20
• Subtask 6: 0 / 15