제출 #1138853

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1138853		dadas08	Election (BOI18_election)	C++20	100 / 100	1062 ms	82180 KiB

election

#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
using namespace std;

using ll = long long;

const ll N = 500'000 + 10;

struct LazySegTree {
	struct Node {
		ll val = -1'000'000'000;
		ll lazy = 0;
		Node() {};
		Node(ll _val, ll _lazy): val(_val), lazy(_lazy) {};
		Node operator+(const Node& other) const {
			Node res;
			res.val = max(this->val, other.val);
			return res;
		}
	};

	ll n_node = 0;
	vector<Node> tree = {};

	void resize(ll _n) { n_node = _n; tree.resize((n_node << 2) + 10); };
	LazySegTree() {};
	LazySegTree(ll _n) { this->resize(_n); };

	void down(ll node, ll l, ll r) {
		tree[node].val += tree[node].lazy;
		if (l < r) {
			tree[node << 1].lazy += tree[node].lazy;
			tree[node << 1 | 1].lazy += tree[node].lazy;
		}
		tree[node].lazy = 0;
	}

	void update(ll node, ll l, ll r, ll L, ll R, ll val) {
		down(node, l, r);
		if (l > r || L > R || l > R || L > r) return;
		if (L <= l && r <= R) {
			tree[node].lazy += val;
			down(node, l, r);
			return;
		}
		ll mid = (l + r) >> 1;
		update(node << 1, l, mid, L, R, val);
		update(node << 1 | 1, mid + 1, r, L, R, val);
		tree[node] = tree[node << 1] + tree[node << 1 | 1];
	}

	void Update(ll l, ll r, ll val) {
		return update(1, 1, n_node, l, r, val);
	}

	Node query(ll node, ll l, ll r, ll L, ll R) {
		down(node, l, r);
		if (l > r || L > R || l > R || L > r) return Node();
		if (L <= l && r <= R) return tree[node];
		ll mid = (l + r) >> 1;
		return query(node << 1, l, mid, L, R) + query(node << 1 | 1, mid + 1, r, L, R);
	}

	Node Query(ll l, ll r) {
		return query(1, 1, n_node, l, r);
	}
};

struct FenWick {
	ll n_node = 0;
	vector<ll> tree = {};

	void resize(ll n) { n_node = n; tree.resize(n + 10, 0); }
	FenWick() {};
	FenWick(ll n) { this -> resize(n); };

	void Update(ll idx, ll val) {
		for( ; idx <= n_node; idx += (idx & - idx)) tree[idx] += val;
		return;
	}

	ll Query(ll idx) {
		ll res = 0;
		for( ; idx > 0; idx -= (idx & - idx)) res += tree[idx];
		return res;
	}

	ll Sum_to(ll l, ll r) {
		if(l > r) return 0;
		return Query(r) - Query(l - 1);
	}

	ll kth(ll val) {
		ll res = 0, s = 0;
		for(ll mask = 30; mask >= 0; -- mask) {
			ll nxt = res | (1 << mask);
			if(nxt < n_node && s + tree[nxt] < val) {
				res = nxt;
				s += tree[nxt];
			}
		}
		++ res; return res;
	}

	void reset(ll idx) {
		for ( ; idx <= n_node; idx += (idx & - idx)) tree[idx] = 0;
		return;
	}
};

ll n, q;
string s;

ll pred[N];
vector<pair<ll, ll>> events[N];
ll lft[N];

ll res[N];

void solve() {
	cin >> n >> s; s = ' ' + s;
	for (ll i = 1; i <= n; ++i) {
		pred[i] = pred[i - 1] + (s[i] == 'C' ? 1 : -1);
	}
	stack<ll> stk;
	for (ll i = n; i >= 1; --i) {
		if (s[i] == 'T') {
			stk.push(i);
		} else {
			if (!stk.empty()) {
				lft[stk.top()] = i;
				stk.pop();
			}
		}
	}
	while (!stk.empty()) {
		stk.pop();
	}
	LazySegTree st(n);
	for (ll i = 1; i <= n; ++i) {
		st.Update(i, i, 1'000'000'000);
		st.Update(i, i, pred[i]);
	}
	cin >> q;
	for (ll i = 1; i <= n; ++i) {
		if (lft[i] != 0) {
			events[lft[i]].push_back(make_pair(i, -1));
		}
	}
	for (ll i = 1; i <= q; ++i) {
		ll l, r; cin >> l >> r;
		events[l].push_back(make_pair(r, i));
	}
	FenWick bit(n);
	for (ll l = n; l >= 1; --l) {
		if (s[l] == 'T') {
			st.Update(l, n, 1);
			bit.Update(l, 1);
		}

		for (const auto& [r, idx] : events[l]) {
			if (idx == -1) {
				st.Update(r, n, -1);
				bit.Update(r, -1);
			} else {
				ll max_pref = max(pred[l - 1], st.Query(l, r - 1).val);
				// cerr << idx << ' ' << max_pref << ' ' << st.Query(r, r).val << '\n';
				res[idx] = bit.Sum_to(l, r) + max(0ll, max_pref - st.Query(r, r).val);
			}
		}
	}
	for (ll i = 1; i <= q; ++i) {
		cout << res[i] << '\n';
	}
}

int32_t main() {
	std::ios_base::sync_with_stdio(false);
	std::cin.tie(nullptr); std::cout.tie(nullptr);
	// ll t; cin >> t;
	ll t = 1;
	while (t-- > 0) {
		solve();
		// cout << '\n';
	}
	return (0 ^ 0);
}

/// Code by vuavisao

• Subtask 1: 28 / 28
• Subtask 2: 54 / 54
• Subtask 3: 18 / 18