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 $\mathrm{nn bracket\; sequences s1,s2,\dots ,sns1,s2,\dots ,sn.\; Calculate\; the\; number\; of\; pairs i,j(1\le i,j\le n)i,j(1\le i,j\le n) such\; that\; the\; bracket\; sequence si+sjsi+sj is\; a\; regular\; bracket\; sequence.\; Operation ++ means\; concatenation\; i.e.\; "()("\; +\; ")()"\; =\; "()()()".}$

If ${\mathrm{si+sjsi+sj and sj+sisj+si are\; regular\; bracket\; sequences\; and i\ne ji\ne j,\; then\; both\; pairs (i,j)(i,j) and (j,i)(j,i) must\; be\; counted\; in\; the\; answer.\; Also,\; ifsi+sisi+si is\; a\; regular\; bracket\; sequence,\; the\; pair (i,i)(i,i) must\; be\; counted\; in\; the\; answer.}}^{}$

Input

The first line contains one integer $\mathrm{n(1\le n\le 3\cdot 105)n(1\le n\le 3\cdot 105) —\; the\; number\; of\; bracket\; sequences.\; The\; following nn lines\; contain\; bracket\; sequences\; — non-empty strings\; consisting\; only\; of\; characters\; "("\; and\; ")".\; The\; sum\; of\; lengths\; of\; all\; bracket\; sequences\; does\; not\; exceed 3\cdot 1053\cdot 105.}$

Output

In the single line print a single integer — the number of pairs $\mathrm{i,j(1\le i,j\le n)i,j(1\le i,j\le n) such\; that\; the\; bracket\; sequence si+sjsi+sj is\; a\; regular\; bracket\; sequence.}$

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 $(1,1),(1,2),(2,1),(2,2)(1,1),(1,2),(2,1),(2,2).$

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#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 + 10;
const int mod = 1e9 + 7;
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, 0, sizeof(le) );
        memset( ri, 0, sizeof(ri) );
        memset( vis, 0, sizeof(vis) );
        for( ll i = 0; i < n; i ++ ) {
            cin >> s[i];
            ll l = 0, r = 0;
            for( ll j = 0; j < s[i].length(); j ++ ) {
                if( s[i][j] == '(' ) {
                    l ++;
                } else if( s[i][j] == ')' && l > 0 ) {
                    l --;
                } else if( s[i][j] == ')' && l == 0 ) {
                    r ++;
                }
            }
            le[i] = l, ri[i] = r;
        }
        for( ll i = 0; i < n; i ++ ) {
            if( le[i] && ri[i] ) {
                vis[i] = 1;
            }
        }
        for( ll i = 0; i < n; i ++ ) {
            if( !vis[i] ) {
                mm[make_pair(le[i],ri[i])] ++;
            }
        }
        ll ans = 0;
        for( ll i = 0; i < n; i ++ ) {
            if( !vis[i] && !ri[i] ) {
                ans += mm[make_pair(0,le[i])];
            }
        }
        cout << ans << endl;
    }
    return 0;
}