Bracket Sequences Concatenation Problem括号序列拼接问题（栈+map+思维）

A bracket(括号) 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 n bracket sequences s1,s2,…,sn. Calculate the number of pairs i,j(1≤i,j≤n) such that the bracket sequence si+sj is a regular bracket sequence. Operation + means concatenation i.e. "()(" + ")()" = "()()()".
If si+sj and sj+si are regular bracket sequences and i≠j, then both pairs (i,j) and (j,i) must be counted in the answer. Also, if si+si is a regular bracket sequence, the pair (i,i) must be counted in the answer.
Input

The first line contains one integer n(1≤n≤3⋅105)— the number of bracket sequences. The following n lines contain bracket sequences — non-empty strings consisting only of characters "(" and ")". The sum of lengths of all bracket sequences does not exceed 3⋅105
.
Output
In the single line print a single integer — the number of pairs i,j(1≤i,j≤n)
such that the bracket sequence si+sj
is a regular bracket sequence.
Examples

Input
3
)
()
(

Output
2
Input
2
()
()

Output
4

Note

In the first example, suitable pairs are (3,1)and (2,2)
.
In the second example, any pair is suitable, namely (1,1),(1,2),(2,1),(2,2)
.

题目意思：有n个字符串，每个字符串都只有'('和')'组成，从中找出两个字符串si,sj( i ! = j)可以构成完全匹配的个数，同样如果si自身也能完全匹配也要算进去。

解题思路：所有的字符串可以分为3类:

1:  自身完美匹配型（即左括号和右括号完美匹配）

2：除去完全匹配的子串，剩下的都是左括号。

3：除去完全匹配的子串，剩下的都是右括号。

对于第一类他的个数ans=c(n,2)*A(2,2)+n(它自身构成的完美匹配），对于第二类和第3类，用map查询一遍（如果有左括号的个数等于右括号的个数，ans=(左括号的种类*右括号的种类），最后不要忘记除去2，因为我们算了两遍。

#include<cstdio>
#include<cstring>
#include<map>
#include<stack>
#include<algorithm>
#define ll long long int
#define MAX 300010
using namespace std;
map<ll,ll>mp;
char str[MAX];
int main()
{
    ll i,n,len,m,k;
    ll counts,ans,sum;
    scanf("%lld",&n);
    getchar();
    m=0;
    ans=0;
    while(n--)
    {
        stack<char>s;
        scanf("%s",str);
        len=strlen(str);
        for(i=0; i<len; i++)
        {
            if(!s.empty())
            {
                if(s.top()=='('&&str[i]==')')
                {
                    s.pop();
                }
                else
                {
                    s.push(str[i]);
                }
            }
            else
            {
                s.push(str[i]);
            }
        }
        if(s.empty())///自身完全匹配
        {
            m++;
        }
        else
        {
            counts=s.size();
            sum=0;
            while(!s.empty())
            {
                if(s.top()=='(')
                {
                    sum++;///记录左括号个数
                }
                s.pop();
            }
            if(sum==0)///剩下的都是右括号
            {
                mp[-counts]++;///负数代表右括号
            }
            else if(sum==counts)///栈里剩下的都是左括号
            {
                mp[counts]++;///正数代表左括号
            }
        }
    }
    map<ll,ll>::iterator it;
    for(it=mp.begin(); it!=mp.end(); it++)
    {
        k=it->first;
        if(mp.count(-k))///只有存在左括号数等于右括号数的才存在完美匹配
        {
            ans+=(ll)(it->second)*mp[-k];
        }
    }
    printf("%lld
",ans/2+m*m);
    return 0;
}