D. Invertible Bracket Sequences
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example:
- bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)");
- bracket sequences ")(", "(" and ")" are not.
Let's define the inverse of the bracket sequence as follows: replace all brackets '(' with ')', and vice versa (all brackets ')' with '('). For example, strings "()((" and ")())" are inverses of each other.
You are given a regular bracket sequence $s$. Calculate the number of pairs of integers $(l,r)$ ($1 \le l \le r \le |s|$) such that if you replace the substring of $s$ from the $l$-th character to the $r$-th character (inclusive) with its inverse, $s$ will still be a regular bracket sequence.
Input
The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.
The only line of each test case contains a non-empty regular bracket sequence; it consists only of characters '(' and/or ')'.
Additional constraint on the input: the total length of the regular bracket sequences over all test cases doesn't exceed $2 \cdot 10^5$.
Output
For each test case, print a single integer — the number of pairs $(l,r)$ meeting the conditions from the statement.
Example
input
4
(())
()
()()()
(()())(())output
1
0
3
13Note
In the first example, there is only one pair:
- $(2, 3)$: (()) $\rightarrow$ ()().
In the second example, there are no pairs.
In the third example, there are three pairs:
- $(2, 3)$: ()()() $\rightarrow$ (())();
- $(4, 5)$: ()()() $\rightarrow$ ()(());
- $(2, 5)$: ()()() $\rightarrow$ (()());
解題思路
先給出我的做法，應該還有更簡潔的方法，等官方題解出了再補。
如果一個括號序列是合法的，那麼必然滿足以下 $2$ 個條件：
- 在任意一個字首中
(的數量不小於)的數量。 - 整個序列的
(和)數量相等。
現在把 ( 和 ) 分別看作 $1$ 和 $-1$，對序列求一個字首和，記為 $s_i$。考慮列舉反轉區間的左端點 $l$，那麼哪些右端點 $r \, (i \leq r \leq n)$ 使得將區間 $[l,r]$ 內的括號反轉後，整個括號序列仍是合法的？
先滿足第 $1$ 個條件。如果反轉區間 $[l,r]$，那麼字首會受到影響的位置就是 $k \in [l,r]$，下標為 $k$ 的字首會變成 $s_{l-1} - (s_{k} - s_{l-1}) = 2 \cdot s_{l-1} - s_k$。如果第條件 $1$ 要滿足，那麼對於每個 $k$ 都必須有 $2 \cdot s_{l-1} - s_k \geq 0 \Rightarrow s_k \leq 2 \cdot s_{l-1}$，即 $\max\limits_{l \leq k \leq r}\{ s_k \} \leq 2 \cdot s_{l-1}$。為此當固定 $l$ 後，可以在區間 $[l,n]$ 內二分出最遠且合法的右端點 $r$，check 的時候需要快速知道某個區間內 $s_i$ 的最大值，這個可以用 RMQ 或線段樹來維護。
最後是滿足第 $2$ 個條件。現在我們確定了最遠的位置 $r$，但並不是所有位於 $[l,r]$ 區間內的下標都適合作為反轉區間的右端點。顯然反轉區間內的 ( 和 ) 的數量必須相同，因此合法的右端點 $k \in [l,r]$ 還需要滿足 $s_k - s_{l-1} = 0$。所以我們現在需要統計在 $l \leq k \leq r$ 內有多少個位置滿足 $s_k = s_{l-1}$，該結果就是以 $l$ 為左端點的合法反轉區間數量。方法很簡單，只需在預處理 $s_i$ 時開個雜湊表儲存每個字首值對應的下標。然後在字首為 $s_{l-1}$ 的下標中二分出大於等於 $l$ 的最小位置 $x$，以及小於等於 $r$ 的最大位置 $y$，合法的右端點數量就是 $y-x+1$。
AC 程式碼如下，時間複雜度為 $O(n \log n)$：
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 2e5 + 5;
char str[N];
int s[N];
int f[18][N];
int query(int l, int r) {
int t = __lg(r - l + 1);
return max(f[t][l], f[t][r - (1 << t) + 1]);
}
void solve() {
scanf("%s", str + 1);
int n = strlen(str + 1);
map<int, vector<int>> mp;
for (int i = 1; i <= n; i++) {
s[i] = s[i - 1] + (str[i] == '(' ? 1 : -1);
mp[s[i]].push_back(i);
}
for (int i = 0; 1 << i <= n; i++) {
for (int j = 1; j + (1 << i) - 1 <= n; j++) {
if (!i) f[i][j] = s[j];
else f[i][j] = max(f[i - 1][j], f[i - 1][j + (1 << i - 1)]);
}
}
LL ret = 0;
for (int i = 1; i <= n; i++) {
int l = i, r = n;
while (l < r) {
int mid = l + r + 1 >> 1;
if (query(i, mid) <= s[i - 1] << 1) l = mid;
else r = mid - 1;
}
if (query(i, l) <= s[i - 1] << 1) {
int x = lower_bound(mp[s[i - 1]].begin(), mp[s[i - 1]].end(), i) - mp[s[i - 1]].begin();
int y = upper_bound(mp[s[i - 1]].begin(), mp[s[i - 1]].end(), l) - mp[s[i - 1]].begin() - 1;
if (x <= y) ret += y - x + 1;
}
}
printf("%lld\n", ret);
}
int main() {
int t;
scanf("%d", &t);
while (t--) {
solve();
}
return 0;
}
參考資料
Educational Codeforces Round 166 [Rated for Div. 2]：https://codeforces.com/blog/entry/129909