Codeforces 990C Bracket Sequences Concatenation Problem

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最新推荐文章于 2022-07-23 10:02:01 发布

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本文介绍了一种算法，用于解决给定多个只包含括号的字符串时，计算能够组合成有效括号序列的不同配对数量的问题。通过预处理字符串来消除相邻匹配的括号对，将问题简化为四种基本情形，并利用哈希映射来统计不同类型的字符串数量。

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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 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

Input

2
()
()

Output

记录一下：卡我初始化。（将s数组开始初始化为0，就一直时间超限）

从n个只有 ‘（’ 和 ‘）’ 的字符串中选出两个，问能组成多少种符合规范的括号，

只有可能是两个非完整括号或两个完整括号组合

思路：先用栈处理一下字符串，相临的 '(' 和 ‘）’ 抵消掉，这种（）不影响判断是否能和其他串组合成合法字符串。

处理后有四种情况：
1，只有左括号，

2，只有右括号，

3，既有左括号，又有右括号，但左括号在右边，右括号在左边，如：）（

4，什么都没有

1,2种情况，可以用map记录，左括号数为负数，右括号数为正数，mp【i】，右括号有i个的字符串的个数

3, 直接统计此类型的个数

4，作废

该非完整括号的字符串个数乘能与之配对的字符串的个数 + 完整括号的个数的平方 $\binom{n}{2}*{A{__2}^{_2}}+n = n^{2}$

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <deque>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define eps 1e-8
#define PI acos(-1)
#define INF 0x3f3f3f3f
#define N 500000 + 10
using namespace std;
const int intN = 3e5 + 10 ;
typedef long long int LL;
const int dir[4][2]= { {1,0},{0,1},{-1,0},{0,-1} };

int GCD(int a,int b)
{
    return b ? GCD(b,a%b) : a;
}
int cmp(int a,int b)
{
    return a>b;
}

int main()
{
    stack<char>p;
    map<int,LL>mp;
    map<int,LL>::iterator it;
    LL i,m=0,t,j,ans=0;
    scanf("%lld",&t);
    getchar();
    for(j=0; j<t; j++)
    {
        char s[N];
        gets(s);
        for(i=0; s[i]; i++)
        {
            if(!p.empty())
            {
                if(p.top()=='(' && s[i]==')')
                    p.pop();
                else
                    p.push(s[i]);
            }
            else
                p.push(s[i]);
        }
        if(p.empty())
            m++;
        else
        {
            int lchar=0,rchar=0;
            while(!p.empty())
            {
                if(p.top()=='(')
                    lchar++;
                else if(p.top()==')')
                    rchar++;
                p.pop();
            }
            if(lchar!=0 && rchar==0)
                mp[-lchar]++;
            else if(lchar==0 && rchar!=0)
                mp[rchar]++;
        }
    }
    for(it=mp.begin(); it!=mp.end(); it++)
    {
        if(mp.find( (it->first)*(-1) )!=mp.end() )
        {
            ans+=it->second*mp[it->first*-1];
            mp[it->first*-1]=0;
        }
    }
    printf("%lld\n",ans+m*m);
    return 0;
}