Submission #828092

#	Time	Username	Problem	Language	Result	Execution time	Memory
828092		Chal1shkan	Arranging Shoes (IOI19_shoes)	C++14	0 / 100	7 ms	9976 KiB

shoes

//Bismillahir-Rahmanir-Rahim

# include <shoes.h>
# include <bits/stdc++.h>

# define pb push_back
# define ff first
# define ss second
# define nl "\n"
# define sz(x) ((int)(x).size())
# define all(x) (x).begin(), (x).end()
# define deb(x) cerr << #x  << " = " << x << endl; 
# define pll pair <ll, ll>
# define pii pair <int, int>
 
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
 
const ll maxn = 1e5 + 7;
const ll inf = 2e18 + 0;
const ll mod = 1e9 + 7;
const ll dx[] = {-1, 1, 0, 0};
const ll dy[] = {0, 0, -1, 1};

using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

ll ans = 0;
pll p[maxn];
vector <ll> qwe[2][maxn];

void upd (ll x, ll id) {
	if (x < 0) {
		p[-x].ff = id;
	}
	else {
		p[x].ss = id;
	}
}

void f (vector <int>& v, ll x1, ll x2) {
	ll dx = 0;
	if (x1 <= x2) {
		dx = 1;
	}
	else {
		dx = -1;
	}
	while (x1 != x2) {
		ans++;
		swap(v[x1], v[x1 + dx]);
		upd(v[x1], x1);
		upd(v[x1 + dx], x1 + dx);
		x1 += dx;
	}
}

ll count (ll x, ll j) {
	ll pos1 = 2 * j, pos2 = 2 * j + 1;
	if (p[x].ff < p[x].ss) {
		return abs(pos1 - p[x].ff) + abs(pos2 - p[x].ss);
	}
	else {
		return abs(pos1 - p[x].ff) + abs(pos2 - p[x].ss) + 1;
	}
}

ll count_swaps (vector <int> S) {
	ll n = sz(S);
	set <ll> st, zxc;
	for (ll i = 0; i < n + n; ++i) {
		zxc.insert(abs(S[i]));
	//	upd(x, i);
		if (S[i] < 0) {
			qwe[0][-S[i]].pb(i);
		}
		else {
			qwe[1][S[i]].pb(i);
		}
	}
	ll sz = 0;
	for (ll x : zxc) {
		for (ll i = 0; i < sz(qwe[0][x]); ++i) {
			++sz;
			st.insert(sz);
			ll pos1 = qwe[0][x][i], pos2 = qwe[1][x][i];
			upd(-sz, pos1);
			upd(sz, pos2);
			S[pos1] = -sz, S[pos2] = sz;
		}
	}
	for (ll j = 0; j < n; ++j) {
		ll best = inf;
		ll opt = -1;
		for (ll x : st) {
			ll dist = count(x, j);
			if (dist < best) {
				best = dist;
				opt = x;
			}
		}
		cout << opt << ' ' << best << nl;
		st.erase(opt);
		f(S, p[opt].ff, 2 * j);
		f(S, p[opt].ss, 2 * j + 1);
	}
	return ans;
}

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