D. Invertible Bracket Sequences

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
13

Note

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:

• $(2, 3)$: ()()() $\rightarrow$ (())();
• $(4, 5)$: ()()() $\rightarrow$ ()(());
• $(2, 5)$: ()()() $\rightarrow$ (()());

 

解题思路

  如果一个括号序列是合法的，那么必然满足以下 $2$ 个条件：

1. 在任意一个前缀中 ( 的数量不小于 ) 的数量。
2. 整个序列的 ( 和 ) 数量相等。

  现在把 ( 和 ) 分别看作 $1$ 和 $-1$，对序列求一个前缀和，记为 $s_i$。考虑枚举反转区间的左端点 $l$，那么哪些右端点 $r \, (i \leq r \leq n)$ 使得将区间 $[l,r]$ 内的括号反转后，整个括号序列仍是合法的？

  先满足第 $1$ 个条件。如果反转区间 $[l,r]$，那么前缀会受到影响的位置就是 $k \in [l,r]$，下标为 $k$ 的前缀会变成 $s_{l-1} - (s_{k} - s_{l-1}) = 2 \cdot s_{l-1} - s_k$。如果第条件 $1$ 要满足，那么对于每个 $k$ 都必须有 $2 \cdot s_{l-1} - s_k \geq 0 \Rightarrow s_k \leq 2 \cdot s_{l-1}$，即 $\max\limits_{l \leq k \leq r}\{ s_k \} \leq 2 \cdot s_{l-1}$。为此当固定 $l$ 后，可以在区间 $[l,n]$ 内二分出最远且合法的右端点 $r$，check 的时候需要快速知道某个区间内 $s_i$ 的最大值，这个可以用 RMQ 或线段树来维护。

  最后是满足第 $2$ 个条件。现在我们确定了最远的位置 $r$，但并不是所有位于 $[l,r]$ 区间内的下标都适合作为反转区间的右端点。显然反转区间内的 ( 和 ) 的数量必须相同，因此合法的右端点 $k \in [l,r]$ 还需要满足 $s_k - s_{l-1} = 0$。所以我们现在需要统计在 $l \leq k \leq r$ 内有多少个位置满足 $s_k = s_{l-1}$，该结果就是以 $l$ 为左端点的合法反转区间数量。方法很简单，只需在预处理 $s_i$ 时开个哈希表存储每个前缀值对应的下标。然后在前缀为 $s_{l-1}$ 的下标中二分出大于等于 $l$ 的最小位置 $x$，以及小于等于 $r$ 的最大位置 $y$，合法的右端点数量就是 $y-x+1$。

  AC 代码如下，时间复杂度为 $O(n \log n)$：

#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;
}

  官方题解的做法与上面的思路一致，不过枚举的是反转区间的右端点 $r$。开一个哈希表记录 $s_i$ 出现的次数，那么以 $r$ 为右端点的合法区间数量就是哈希表中值 $s_r$ 的数量。接着把 $s_r$ 插入到哈希表中，为了使得哈希表中的 $s_i$ 都满足第 $1$ 个条件（在插入 $s_r$ 前就已经满足了），需要将哈希表中使得 $s_i < s_r - s_i$ 的 $s_i$ 删除。

  AC 代码如下，时间复杂度为 $O(n \log{n})$：

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N = 2e5 + 5;

char s[N];

void solve() {
    scanf("%s", s + 1);
    LL ret = 0;
    map<int, int> mp;
    for (int i = 1, sum = 0; s[i]; i++) {
        sum += s[i] == '(' ? 1 : -1;
        ret += mp[sum];
        while (!mp.empty() && mp.begin()->first * 2 < sum) {
            mp.erase(mp.begin());
        }
        mp[sum]++;
    }
    printf("%lld\n", ret);
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        solve();
    }
    
    return 0;
}

 

参考资料

  Educational Codeforces Round 166 Editorial：https://codeforces.com/blog/entry/129992