Submission #247043

#	Time	Username	Problem	Language	Result	Execution time	Memory
247043		ernestvw	Arranging Shoes (IOI19_shoes)	C++17	100 / 100	395 ms	288632 KiB

shoes

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

using ll = long long;
#define sz(x) ((int)x.size())

ll tree[1000*1000];
ll lazy[1000*1000];

void update(int l, int r, int L, int R, int i, ll val) {
	if(r < l || l > R || r < L) return;
	if(L <= l && r <= R) {
		tree[i] += lazy[i] + val;
		lazy[2*i+1] += lazy[i] + val;
		lazy[2*i+2] += lazy[i] + val;
		lazy[i] = 0;
		return;
	}
	lazy[2*i+1] += lazy[i];
	lazy[2*i+2] += lazy[i];
	lazy[i] = 0;
	update(l, (l+r)/2, L, R, 2*i+1, val);
	update((l+r)/2+1, r, L, R, 2*i+2, val);
	tree[i] = tree[2*i+1] + tree[2*i+2];
}

ll sum(int l, int r, int L, int R, int i) {
	if(r < l || l > R || r < L) return 0;
	if(L <= l && r <= R) {
		tree[i] += lazy[i];
		lazy[2*i+1] += lazy[i];
		lazy[2*i+2] += lazy[i];
		lazy[i] = 0;
		return tree[i];
	}
	lazy[2*i+1] += lazy[i];
	lazy[2*i+2] += lazy[i];
	tree[i] += lazy[i];
	lazy[i] = 0;
	return sum(l, (l+r)/2, L, R, 2*i+1) + sum((l+r)/2+1, r, L, R, 2*i+2);
}

ll count_swaps(vector<int> S) {
	int n = (int)S.size();
	ll ans = 0;

	vector<deque<int>> L(n+1), R(n+1);
//	vector<ll> avant(n+1,0);
//	for(int i = 0; i < n; ++i)
//		avant[i] = n - i - 1;
	
	memset(tree, 0, sizeof(tree));
	for(int i = 0; i < n; ++i)
		update(0, n-1, i, i, 0, n-i-1);

	for(int i = 0; i < n; ++i) {
//		cout << "----" << endl;
//		for(int i = 0; i < n; ++i) cout << avant[i] << " ";
//		cout << endl;
//		for(int i = 0; i < n; ++i) cout << sum(0, n-1, i, i, 0) << " ";
//		cout << endl;
		
		int x = abs(S[i]);
		if(S[i] > 0) {
			if(sz(L[x]) > 0) {
				int j = L[x].front();
//				ans += avant[j] - avant[i] - 1LL;
//				for(int k = j + 1; k <= i; ++k)
//					avant[k]--;
				ans += sum(0, n-1, j, j, 0) - sum(0, n-1, i, i, 0) - 1LL;
				update(0, n-1, j+1, i, 0, -1);
				L[x].pop_front();
			}
			else R[x].push_back(i);
		}
		else {
			if(sz(R[x]) > 0) {
				int j = R[x].front();
//				ans += avant[j] - avant[i];
//				for(int k = j; k <= i; ++k)
//					avant[k]--;
				ans += sum(0, n-1, j, j, 0) - sum(0, n-1, i, i, 0);
				update(0, n-1, j, i, 0, -1);
				R[x].pop_front();
			}
			else L[x].push_back(i);
		}
	}

	return ans;
}

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