div2 #603 括號專題+線段樹維護區間（1~n）和的最大值和最小值和本身

題目鏈接
題目大意：
我們定義合法的字符串爲 T, (T), ((T)(T))… 他給出n個操作，RL代表左移，剩下的字符就表示把當前位置上的字符替換爲現在的字符。現在有個染色規則，如果兩個括號是內嵌（nested）的關係，那他們的顏色就一定要不一樣，然後輸出每步操作後染色的最少顏色，如果不是合法字符串， 那就輸出-1.

仔細一想，最少顏色其實就是字符串的最大深度。我們可以用線段樹維護1~1e6區間內的區間和的最大值和最小值，如果區間和不爲0或者最小值 < 0，那就不是合法字符串，輸出-1

ac代碼

#include <bits/stdc++.h>

#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)
#define rep(i, l, r) for (int i = (l); i <= (r); i++)
#define per(i, r, l) for (int i = (r); i >= (l); i--)
#define lson node<<1
#define rson node<<1|1
 
using namespace std;
 
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector<vi> vvi;
typedef long long i64;
typedef vector<i64> vi64;
typedef vector<vi64> vvi64;
typedef pair<i64, i64> pi64;
typedef double ld;
 
template<class T> bool uin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool uax(T &a, T b) { return a < b ? (a = b, true) : false; }

const int maxn = (int)1e6 + 1000;
const int N = (int)1e6 + 100;
int n, sum[maxn << 2], mi[maxn << 2], ma[maxn << 2];
string s;
char x[maxn];
 
void build(int l, int r, int node) {
	sum[node] = mi[node] = ma[node] = 0;
	if (l == r) return;
	int mid = (l + r) >> 1;
	build(l, mid, lson);
	build(mid + 1, r, rson);
}
void pushup(int node) {
	sum[node] = sum[lson] + sum[rson];
	mi[node] = min(mi[lson], sum[lson] + mi[rson]);
	ma[node] = max(ma[lson], sum[lson] + ma[rson]);
}
void update(int l, int r, int node, int pos, int num) {
	if (l == r) {
		sum[node] = mi[node] = ma[node] = num;
		return;
	}
	int mid = (l + r) >> 1;
	if (pos <= mid) update(l, mid, lson, pos, num);
	else update(mid + 1, r, rson, pos, num);
	pushup(node);
}

int main() {
	ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.precision(10);
    cout << fixed;
#ifdef LOCAL_DEFINE
    freopen("in", "r", stdin);
#endif
	
	cin >> n;
	cin >> s;
	int now = 1;
	build(1, N, 1);
	forn(i, n) {
		if (s[i] == 'R') ++now;
		else if (s[i] == 'L') {
			if (now > 1) --now;
		} else if (s[i] == '(') {
			update(1, N, 1, now, 1);
			x[now] = '(';
		} else if (s[i] == ')') {
			update(1, N, 1, now, -1);
			x[now] = ')';
		} else {
			char ch = x[now];
			update(1, N, 1, now, 0);
		}
		if (sum[1] != 0 || mi[1] < 0) cout << -1 << " ";
		else cout << ma[1] << " ";
	}

#ifdef LOCAL_DEFINE
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
	return 0;
}