Codeforces Round #603 (Div. 2) E. Editor（线段树）

链接：

https://codeforces.com/contest/1263/problem/E

题意：

The development of a text editor is a hard problem. You need to implement an extra module for brackets coloring in text.

Your editor consists of a line with infinite length and cursor, which points to the current character. Please note that it points to only one of the characters (and not between a pair of characters). Thus, it points to an index character. The user can move the cursor left or right one position. If the cursor is already at the first (leftmost) position, then it does not move left.

Initially, the cursor is in the first (leftmost) character.

Also, the user can write a letter or brackets (either (, or )) to the position that the cursor is currently pointing at. A new character always overwrites the old value at that position.

Your editor must check, whether the current line is the correct text. Text is correct if the brackets in them form the correct bracket sequence.

Formally, correct text (CT) must satisfy the following rules:

any line without brackets is CT (the line can contain whitespaces);
If the first character of the string — is (, the last — is ), and all the rest form a CT, then the whole line is a CT;
two consecutively written CT is also CT.
Examples of correct texts: hello(codeforces), round, ((i)(write))edi(tor)s, ( me). Examples of incorrect texts: hello)oops(, round), ((me).

The user uses special commands to work with your editor. Each command has its symbol, which must be written to execute this command.

The correspondence of commands and characters is as follows:

L — move the cursor one character to the left (remains in place if it already points to the first character);
R — move the cursor one character to the right;
any lowercase Latin letter or bracket (( or )) — write the entered character to the position where the cursor is now.
For a complete understanding, take a look at the first example and its illustrations in the note below.

You are given a string containing the characters that the user entered. For the brackets coloring module's work, after each command you need to:

check if the current text in the editor is a correct text;
if it is, print the least number of colors that required, to color all brackets.
If two pairs of brackets are nested (the first in the second or vice versa), then these pairs of brackets should be painted in different colors. If two pairs of brackets are not nested, then they can be painted in different or the same colors. For example, for the bracket sequence ()(())()() the least number of colors is 2, and for the bracket sequence (()(()())())(()) — is 3.

Write a program that prints the minimal number of colors after processing each command.

思路：

先想了线段树，但是不知道怎么维护。
看了题解，发现维护前缀和，前缀和的最大最小值。
最大值判断嵌套深度，最小值没有负数说明满足条件。

代码：

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e6+10;

struct Node
{
    int sum;
    int maxv, minv;
    int lazy;
}node[MAXN*4];

char s[MAXN], e[MAXN];
int n, ans[MAXN];

void Build(int root, int l, int r)
{
    node[root].sum = node[root].maxv = node[root].minv = node[root].lazy = 0;
    if (l == r)
        return ;
    int mid = (l+r)/2;
    Build(root<<1, l, mid);
    Build(root<<1|1, mid+1, r);
}

void PushUp(int root)
{
    node[root].sum = node[root<<1].sum+node[root<<1|1].sum;
    node[root].minv = min(node[root<<1].minv, node[root<<1|1].minv);
    node[root].maxv = max(node[root<<1].maxv, node[root<<1|1].maxv);
}

void PushDown(int root, int l, int r)
{
    if (node[root].lazy != 0)
    {
        int mid = (l+r)/2;
        node[root<<1].lazy += node[root].lazy;
        node[root<<1|1].lazy += node[root].lazy;
        node[root<<1].sum += node[root].lazy*(mid-l+1);
        node[root<<1|1].sum += node[root].lazy*(r-mid);
        node[root<<1].minv += node[root].lazy;
        node[root<<1|1].minv += node[root].lazy;
        node[root<<1].maxv += node[root].lazy;
        node[root<<1|1].maxv += node[root].lazy;
        node[root].lazy = 0;
    }
}

void Update(int root, int l, int r, int ql, int qr, int v)
{
    if (qr < l || r < ql)
        return;
    if (ql <= l && r <= qr)
    {
        node[root].sum += (r-l+1)*v;
        node[root].maxv += v;
        node[root].minv += v;
        node[root].lazy += v;
        return;
    }
    PushDown(root, l, r);
    int mid = (l+r)/2;
    Update(root<<1, l, mid, ql, qr, v);
    Update(root<<1|1, mid+1, r, ql, qr, v);
    PushUp(root);
}

int QuerySum(int root, int l, int r, int p)
{
    if (l == r)
        return node[root].sum;
    int mid = (l+r)/2;
    PushDown(root, l, r);
    if (p <= mid)
        return QuerySum(root<<1, l, mid, p);
    else
        return QuerySum(root<<1|1, mid+1, r, p);
}

int main()
{
    scanf("%d", &n);
    scanf("%s", s);
    Build(1, 1, n);
    int p = 1, len = strlen(s);
    for (int i = 0;i < len;i++)
    {
        if (s[i] == 'L')
        {
            if (p>1)
                p--;
        }
        else if (s[i] == 'R')
            p++;
        else
        {
            if (s[i] == '(' && e[p] == ')')
                Update(1, 1, n, p, n, 2);
            else if (s[i] == ')' && e[p] == '(')
                Update(1, 1, n, p, n, -2);
            else if (s[i] == '(' && e[p] != '(')
                Update(1, 1, n, p, n, 1);
            else if (s[i] == ')' && e[p] != ')')
                Update(1, 1, n, p, n, -1);
            else if (s[i] != '(' && s[i] != ')')
            {
                if (e[p] == '(')
                    Update(1, 1, n, p, n, -1);
                else if (e[p] == ')')
                    Update(1, 1, n, p, n, 1);
            }
            e[p] = s[i];
        }
        int sum = QuerySum(1, 1, n, n);
//        cout << sum << endl;
        int minv = node[1].minv;
        if (sum == 0 && minv >= 0)
            ans[i] = node[1].maxv;
        else
            ans[i] = -1;
    }
    for (int i = 0;i < len;i++)
        cout << ans[i] << ' ';
    cout << endl;

    return 0;
}