CF990C Bracket Sequences Concatenation Problem 思维 第五道 括号经典处理题目

 Bracket Sequences Concatenation Problem

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A bracket sequence is a string containing only characters "(" and ")".

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 "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.

You are given nn bracket sequences s1,s2,…,sns1,s2,…,sn. Calculate the number of pairs i,j(1≤i,j≤n)i,j(1≤i,j≤n) such that the bracket sequence si+sjsi+sj is a regular bracket sequence. Operation ++ means concatenation i.e. "()(" + ")()" = "()()()".

If si+sjsi+sj and sj+sisj+si are regular bracket sequences and i≠ji≠j, then both pairs (i,j)(i,j) and (j,i)(j,i) must be counted in the answer. Also, ifsi+sisi+si is a regular bracket sequence, the pair (i,i)(i,i) must be counted in the answer.

Input

The first line contains one integer n(1≤n≤3⋅105)n(1≤n≤3⋅105) — the number of bracket sequences. The following nn lines contain bracket sequences — non-empty strings consisting only of characters "(" and ")". The sum of lengths of all bracket sequences does not exceed 3⋅1053⋅105.

Output

In the single line print a single integer — the number of pairs i,j(1≤i,j≤n)i,j(1≤i,j≤n) such that the bracket sequence si+sjsi+sj is a regular bracket sequence.

Examples

input

Copy

3
)
()
(

output

Copy

2

input

Copy

2
()
()

output

Copy

4

Note

In the first example, suitable pairs are (3,1)(3,1) and (2,2)(2,2).

In the second example, any pair is suitable, namely (1,1),(1,2),(2,1),(2,2)(1,1),(1,2),(2,1),(2,2).

题意： 给你n个只包含'('和')'的字符串，问每两个字符串相互组合后形成的完整括号（由左到右都可以匹配）的种数

首先消除每个字符串内的已匹配的括号，如果剩余的字符串还同时含有左括号和右括号，那么这种字符串是不可以和其他字符串匹配成功的，直接剔除。

最后枚举只剩余左括号的，看有多少右括号与之对应，这样避免了重复计数。

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e6 + ;
const int mod = 1e9 + ;
typedef long long ll;
string s[maxn];
map< pair< ll, ll >, ll > mm;
ll le[maxn], ri[maxn], vis[maxn];
int main(){
    std::ios::sync_with_stdio(false);
    ll n;
    while( cin >> n ) {
        mm.clear();
        memset( le, , sizeof(le) );
        memset( ri, , sizeof(ri) );
        memset( vis, , sizeof(vis) );
        for( ll i = ; i < n; i ++ ) {
            cin >> s[i];
            ll l = , r = ;
            for( ll j = ; j < s[i].length(); j ++ ) {
                if( s[i][j] == '(' ) {
                    l ++;
                } else if( s[i][j] == ')' && l >  ) {
                    l --;
                } else if( s[i][j] == ')' && l ==  ) {
                    r ++;
                }
            }
            le[i] = l, ri[i] = r;
        }
        for( ll i = ; i < n; i ++ ) {
            if( le[i] && ri[i] ) {
                vis[i] = ;
            }
        }
        for( ll i = ; i < n; i ++ ) {
            if( !vis[i] ) {
                mm[make_pair(le[i],ri[i])] ++;
            }
        }
        ll ans = ;
        for( ll i = ; i < n; i ++ ) {
            if( !vis[i] && !ri[i] ) {
                ans += mm[make_pair(,le[i])];
            }
        }
        cout << ans << endl;
    }
    return ;
}