Codeforces 1263E.Editor

cf 1263E. Editor

Problem Description

传送门

题意

给你一段序列，其中L代表左移，R代表右移，小写字母代表在当前位置放置对应字母，),(代表在当前位置放置左右括号，当括号可以匹配时输出括号的最大层数，否则输出-1

题解

• 小写字母是无意义的操作
• 当（时我们在序列当前位置加1，)我们在当前位置减1。如此最大层数就是序列前缀和的最大值，括号能否匹配取决于前缀和最小的是否等于0以及总共的左右括号数是否相等
• 线段树区间修改，区间最值

C++

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <queue>
#include <set>
using namespace std;

#define ll long long

const int N = 1e6 + 50;

char a[N];
ll maxn[4 * N], sum[4 * N], minn[4 * N], sumf[4 * N], tag[4 * N];
int n, q, z;
int cnt;

void pushup(int o, int l, int r)
{
	int lo = 2 * o, ro = 2 * o + 1, m = (l + r) >> 1;
	sum[o] = sum[lo] + sum[ro];
	maxn[o] = max(maxn[lo], maxn[ro]);
	minn[o] = min(minn[lo], minn[ro]);
	sumf[o] = sumf[lo] + sumf[ro];
}

void build(int o, int l, int r)
{
	if (l == r)
	{
		sum[o] = maxn[o] = minn[o] = 0;
		sumf[o] = 0;
		return;
	}
	int lo = 2 * o, ro = 2 * o + 1, m = (l + r) >> 1;
	build(lo, l, m);
	build(ro, m + 1, r);
	pushup(o, l, r);
}

void pushdown(int o, int l, int r)
{
	if (l == r)
		return;
	int lo = 2 * o, ro = 2 * o + 1, m = (l + r) >> 1;
	if (tag[o])
	{
		tag[lo] += tag[o];
		tag[ro] += tag[o];
		sumf[lo] += (ll)(m - l + 1) * tag[o] * tag[o] + (ll)2 * tag[o] * sum[lo];
		sumf[ro] += (ll)(r - m) * tag[o] * tag[o] + (ll)2 * tag[o] * sum[ro];
		sum[lo] += (ll)(m - l + 1) * tag[o];
		sum[ro] += (ll)(r - m) * tag[o];
		maxn[lo] += tag[o];
		maxn[ro] += tag[o];
		minn[lo] += tag[o];
		minn[ro] += tag[o];
	}
	tag[o] = 0;
}

int addl, addr, addi;
void add(int o, int l, int r)
{
	if (l >= addl && r <= addr)
	{
		sumf[o] += (ll)(r - l + 1) * addi * addi + 2ll * addi * sum[o];
		sum[o] += (ll)(r - l + 1) * addi;
		maxn[o] += addi;
		minn[o] += addi;
		tag[o] += addi;
		return;
	}
	pushdown(o, l, r);
	int lo = 2 * o, ro = 2 * o + 1, m = (l + r) >> 1;
	if (addl <= m)
		add(lo, l, m);
	if (addr > m)
		add(ro, m + 1, r);
	pushup(o, l, r);
}

ll qsum = 0, qsumf = 0;
int ql, qr;
void query(int o, int l, int r)
{
	if (l >= ql && r <= qr)
	{
		qsum += sum[o];
		qsumf += sumf[o];
		return;
	}
	int lo = 2 * o, ro = 2 * o + 1, m = (l + r) >> 1;
	pushdown(o, l, r);
	if (ql <= m)
		query(lo, l, m);
	if (qr > m)
		query(ro, m + 1, r);
	pushup(o, l, r);
}
string s;
int main()
{
#ifdef ONLINE_JUDGE
#else
	freopen("r.txt", "r", stdin);
#endif
	ios::sync_with_stdio(false);
	cin >> n >> s;

	build(1, 1, n);
	for (int i = 1; i <= n; i++)
		a[i] = 'a';

	cin >> n >> s;
	int now = 1;
	for (int i = 0; i < n; i++)
	{
		if (s[i] == 'L')
		{
			now--;
			now = max(now, 1);
		}
		else if (s[i] == 'R')
		{
			now++;
		}
		else
		{
			addl = now;
			addr = n;

			if (a[now] == '(')
			{
				if (s[i] == '(')
					addi = 0;
				else if (s[i] == ')')
					addi = -2;
				else
					addi = -1;
			}
			else if (a[now] == ')')
			{
				if (s[i] == ')')
					addi = 0;
				else if (s[i] == '(')
					addi = 2;
				else
					addi = 1;
			}
			else
			{
				if (s[i] == '(')
					addi = 1;
				else if (s[i] == ')')
					addi = -1;
				else
					addi = 0;
			}
			add(1, 1, n);
			cnt += addi;
			a[now] = s[i];
		}
#ifdef debug
		cout << maxn[1] << " " << minn[1] << endl;
#else
		if (cnt == 0 and minn[1] == 0)
			cout << maxn[1] << " ";
		else
			cout << -1 << " ";
#endif
	}
}