CodeForces 990C

Description

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

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 $\mathrm{i,j(1\le i,j\le n)i,j(1\le i,j\le n) such\; that\; the\; bracket\; sequence si+sjsi+sj is\; a\; regular\; bracket\; sequence.}$

Sample Input

Input

3
)
()
(

Output

2

Input

2
()
()

Output

4

Hint

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

 

括号匹配问题，需要注意几个细节。

有以下几种情况是不能与其他组括号构成合法的括号：)(         )(((      )))(

已经是合法的括号组不需要与不合法的括号组匹配。

 

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<deque>
#include<map>
#include<iostream>
using namespace std;
typedef long long  LL;
const double pi=acos(-1.0);
const double e=exp(1);
const int N = 300010;

#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r

map<LL,LL> mp;
char con[N];
int main()
{
    LL i,p,j,n,check;
    LL cont=0,ans=0,len1,len2;
    scanf("%lld",&n);
    getchar();
    for(j=1;j<=n;j++)
    {
        p=check=0;
        len1=len2=0;
        memset(con,0,sizeof(0));
        scanf("%s",con);
        for(i=0;i<300009;i++)
        {
            if(con[i]==0)
                break;
            if(con[i]=='(')
            {
                len1++;
                p++;
            }
            else
            {
                p--;
                if(len1)
                    len1--;
                else
                    len2++;
            }
        }
        if(len1==0&&len2==0)
            cont++;
        else
        {
            if(len1==0)
                mp[p]++;
            if(len2==0)
                mp[p]++;
        }
    }
    ans=cont*cont;

    map<LL,LL>::iterator it1;
    for(it1=mp.begin();it1!=mp.end();it1++)
    {
        if(it1->first>0)
            break;
        if(mp[-(it1->first)]>0)
            ans+=(it1->second)*mp[-(it1->first)];
    }
    printf("%lld\n",ans);
    return 0;
}