code forces 990C

http://codeforces.com/contest/990/problem/C

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).

 

题意，寻找合法的括号匹配对数（1,3）和（3,1）不同

题解：感谢llw大佬讲题，思路是将每一组括号转化成对应的值来匹配，计算可匹配的值，相加即可得到答案

#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e5+5;
typedef long long ll;
ll n,mx=0;
char str[maxn];
map<ll,ll> mp;
int main(){
    ios::sync_with_stdio(false);
    cin>>n;
    while(n--){
        cin>>str;
        ll flag=0,k=0,len=strlen(str);
        for(int i=0;i<len;i++){
              if(str[i]=='(') k++;
              else k--;
              //k=0的情况是这组数据符合要求
              //k>0的情况是左大于右
              //k<0的情况是右大于左
              if(k<0) flag=min(flag,k);
        }
        if(flag<0&&k>flag) continue;  //不合法的序列不用管,譬如())(
        else{
            //cout<<k<<" "<<flag<<endl;
            mp[k]++;
            mx=max(k,mx);
        }
    }
    ll ans=0;
    for(int i=0;i<=mx;i++){
        ans+=mp[i]*mp[-i];
    }
    cout<<ans<<endl;
    return 0;
}

 

来源：oschina

链接：https://my.oschina.net/u/4340671/blog/3941151