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=;
     ans=;
     while(n--)
     {
         stack<char>s;
         scanf("%s",str);
         len=strlen(str);
         for(i=; 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=;
             while(!s.empty())
             {
                 if(s.top()=='(')
                 {
                     sum++;///记录左括号个数
                 }
                 s.pop();
             }
             if(sum==)///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\n",ans/+m*m);
     return ;
 }