Submission #1345540

#TimeUsernameProblemLanguageResultExecution timeMemory
1345540limitsArranging Shoes (IOI19_shoes)C++20
70 / 100
1100 ms146620 KiB
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3,unroll-loops")

#include <bits/stdc++.h>

using namespace std;

#define f0r(i, n) for (auto i = 0; i < (n); ++i)
#define fnr(i, n, k) for (auto i = (n); i < (k); ++i)
#define all(v) (v).begin(), (v).end()
#define pb push_back
#define F first
#define S second
#define ctn(x) cout << x << '\n'
#define forl(a, l) for (auto a : l)
#define ctl(l) for (auto &a : (l)) cout << a << ' '; cout << endl;
#define lb(v, x) (lower_bound(all(v), x) - begin(v))
#define ub(v, x) (upper_bound(all(v), x) - begin(v))
#define pq priority_queue

template <class T>
using V = vector<T>;
using ll = long long;
using vi = V<int>;
using vl = V<ll>;
using pi = pair<int, int>;
using ti = tuple<int, int, int>;
using Adj = V<vi>;
using vvi = V<vi>;

#include "shoes.h"

ll solve(vi::iterator s, vi::iterator e) {
	if (s == e) return 0;
	int x = *s;
	if (x < 0) {
		auto k = s;
		while (*k != -x) ++k;
		swap(*k, *(s+1));
		return (k - s - 1) + solve(s+2, e);
	} else {
		auto k = s;
		while (*k != -x) ++k;
		swap(*k, *(s+1));
		return (k - s) + solve(s+2, e);
	}
}

ll count_swaps(vi s) {
	int n = s.size()/2;
	if (n == 1) return s[0] > s[1];
	/*
	ll cnt = 0;
	map<int, set<int>> mp;
	f0r(i, 2*n) {
		int x = s[i];
		if (x < 0) {
			if (mp[-x].size()) {
				int y = *mp[-x].begin();
				mp[-x].erase(y);
				cnt += i - y;
			} else mp[x].insert(i);
		} else if (x > 0) {
			if (mp[-x].size()) {
				int y = mp[-x].front();
				mp[-x].pop();
				cnt += i - y - 1;
			} else mp[x].push(i);
		}
	}
	*/
	map<int, queue<int>> mp;
	f0r(i, 2*n) mp[s[i]].push(i);
	ll cnt = 0;
	vi ad(2*n);
	V<bool> dn(2*n);
	for (int i = 0; i < 2*n; i++) if (!dn[i]) {
		int x = s[i], j = mp[-x].front();
		//cout << x << ' ' << i << ' ' << j << endl;
		mp[x].pop(), mp[-x].pop();
		int x2 = i + ad[i], j2 = j + ad[j];
		//cout << x2 << ' ' << j2 << endl;
		cnt += x < 0 ? (j2 - x2 - 1) : (j2-x2);
		for (int y = i; y <= j; ++y) ad[y]++;

		dn[j] = 1;
	}
	return cnt;
}
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