C.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, 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 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).

 

先预处理剩余的左括号和右括号，再用pair和map存，如果剩余的左括号和右括号都不会为零，那么这个串必然不能和其他串组成合法串。最后枚举剩余左括号的，看有多少个对应的右括号，这样不会重。

#include <bits/stdc++.h>
#define maxn 300005
using namespace std;
string a[maxn];
typedef long long ll;
int lefts[maxn]={0};
int rights[maxn]={0};
bool vis[maxn]={0};
map<pair<int,int>,int> mp;
int main()
{
    int n,j,i;
    cin>>n;
    for(i=0;i<n;i++)
    {
        cin>>a[i];
    }
    for(i=0;i<n;i++)
    {   int l=0,r=0;
        for(j=0;j<a[i].size();j++)
        {
            if(a[i][j]=='(')
            {
                l++;
            }
            else if(a[i][j]==')'&&l>0)
            {
                l--;
            }
            else if(a[i][j]==')'&&l==0)
            {
                r++;
            }
        }
        lefts[i]=l;
        rights[i]=r;
    }
    for(i=0;i<n;i++)
    {
        if(lefts[i]!=0&&rights[i]!=0)
        {
            vis[i]=true;
        }
    }
    for(i=0;i<n;i++)
    {
        if(!vis[i])
        {
            mp[make_pair(lefts[i],rights[i])]++;
        }
    }
    ll ans=0;
    for(i=0;i<n;i++)
    {
        if(!vis[i]&&rights[i]==0)
        {
        ans+=(ll)mp[make_pair(0,lefts[i])];
        }
    }
    cout<<ans<<endl;
    return 0;
}

　　

来源：oschina

链接：https://my.oschina.net/u/4359607/blog/3940538