Codeforces 990C （思维）

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题面：

    

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 $n$ bracket sequences $s_1,s_2,\dots ,s_n$. Calculate the number of pairs $i,j(1\le i,j\le n)$ such that the bracket sequence $s_i+s_j$ is a regular bracket sequence. Operation $+$ means concatenation i.e. "()(" + ")()" = "()()()".

If $s_i+s_j$ and $s_j+s_i$ are regular bracket sequences and $i\ne j$, then both pairs $(i,j)$ and $(j,i)$ must be counted in the answer. Also, if $s_i+s_i$ is a regular bracket sequence, the pair $(i,i)$ must be counted in the answer.

Input

The first line contains one integer $n(1\le n\le 3\cdot {10}^5)$ — 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 $3\cdot {10}^5$.

Output

In the single line print a single integer — the number of pairs $i,j(1\le i,j\le n)$ such that the bracket sequence $s_i+s_j$ 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 $(3,1)$ and $(2,2)$.

In the second example, any pair is suitable, namely $(1,1),(1,2),(2,1),(2,2)$.

题目描述：

    给你n个总长度不超过3e5的“）”或“（”的字符串。当存在完整的括号“（）”，则称这是完美的括号。问你从这n个选任意两个字符串并将它们相加，最多能构成多少个完美的括号。

题目分析：

    首先我们可以观察到，对于每一个字符串，我们将从左到右将“（”和“）”括号进行匹配。倘若剩下的括号既有“（”又有“）”，则存在且仅存在“）（”这种字符串。而我们很容易发现，这种字符串是不可能进行匹配的，因此我们只需要排除掉这种情况即可。

    而倘若不存在上述的情况，我们不难发现，此时必定仅剩下一种符号。此时，我们可以开一个桶num[i][j]，使得num[i][j]++,表示j个第i种括号的数量+1。

    统计完这些数据之后，我们只需从头遍历到n个字符串中最大的长度len，每一次将num[0][len]*num[1][len]（假设0代表“（”；1代表“）”）加到结果去即可。

代码：

#include <bits/stdc++.h>
#define maxn 300005
using namespace std;
typedef long long ll;
ll num[2][maxn];
int main()
{
    int n;
    string str;
    cin>>n;
    ll res=0;
    for(int i=0;i<n;i++){
        cin>>str;
        int len=str.length();
        ll l=0,r=0;
        for(int j=0;j<len;j++){
            if(str[j]==')'){
                if(l) l--;//如果匹配到右括号而前面出现过左括号则左右括号匹配
                else r++;//如果匹配不出来，则右括号+1
            }
            else l++;
        }
        if(l&&r) continue;//如果左右括号均有剩余，则只可能出现“）（”的情况
        if(l){
            num[0][l]++;//剩余l个左括号匹配，使桶的数量+1
        }
        else if(r){
            num[1][r]++;
        }
        else res++;//完全匹配
    }
    res*=res;//使得完全匹配能够翻倍
    for(int i=1;i<maxn;i++){//统计0到最大长度的数量
        res+=num[0][i]*num[1][i];
    }
    cout<<res<<endl;
}