Balanced Sequence

Problem Description

Chiaki has nnn strings s1,s2,…,sn" role="presentation" style="position: relative;">s1,s2,…,sns1,s2,…,sns_1,s_2,\dots,s_n consisting of ‘(’ and ‘)’. A string of this type is said to be balanced:

if it is the empty string
if AAA and B" role="presentation" style="position: relative;">BBB are balanced, ABABAB is balanced,
if AAA is balanced, (A)" role="presentation" style="position: relative;">(A)(A)(A) is balanced.

Chiaki can reorder the strings and then concatenate them get a new string ttt. Let f(t)" role="presentation" style="position: relative;">f(t)f(t)f(t) be the length of the longest balanced subsequence (not necessary continuous) of ttt. Chiaki would like to know the maximum value of f(t)" role="presentation" style="position: relative;">f(t)f(t)f(t) for all possible ttt.

Input

There are multiple test cases. The first line of input contains an integer T" role="presentation" style="position: relative;">TTT, indicating the number of test cases. For each test case: The first line contains an integer nnn (1≤n≤105" role="presentation" style="position: relative;">1≤n≤1051≤n≤1051 \le n \le 10^5) – the number of strings. Each of the next nnn lines contains a string si" role="presentation" style="position: relative;">sisis_i (1≤|si|≤1051≤|si|≤1051 \le |s_i| \le 10^5) consisting of (' and)’. It is guaranteed that the sum of all |si||si||s_i| does not exceeds 5×1065×1065 \times 10^6.

Output

For each test case, output an integer denoting the answer.

Sample Input

2
1
)()(()(
2
)
)(

Sample Output

4
2

题目概述

有n行只包含左右括号的字符串，问如何连接这些字符串可以使得匹配到的括号最多（包括本行自身含有的“（）”）

解题思路

我们先算每一行字符串中含有的成对的（），并在该行中删除已配对过的“（” 和 “）”
进行完以上操作我们可以发现每行数组都只剩下多个“）” “（”。且）全在左边，（全在右边。（每个字符串都这么处理后剩下都为“)))((((((”结构的括号串。）
那么我们把剩下的字符串拼接
要想匹配到的括号最多，则尽量让每个括号都起作用了；
如果只包含左括号则把他放在最前边，如果只有右括号把他放在最后边；
可以看出左括号多于右括号的在前，反之在后；
同是左括号多于右括号的，右括号少的在前；
同是右括号多于左括号的，左括号少的在后；

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>using namespace std;
const int mmax=1e5+7;
char st[mmax];
struct node
{int l,r;
}a[mmax];bool cmp(node x,node y)
{if (x.r >= x.l && y.r < y.l)return false;if (x.r < x.l && y.r >= y.l)return true; //')'比'('多的放后面 //')'比'('少   比较')'少的放前面 ')'比'('多   比较'（'少的放后面if (x.r < x.l && y.r < y.l)return x.r<y.r;else return x.l>y.l;
}
int main()
{int t;scanf("%d",&t);while(t--){int n,sum=0;scanf("%d",&n);for(int i=0;i<n;i++){scanf("%s",st);int len=strlen(st);a[i].l=0;a[i].r=0;for(int j=0;j<len;j++){if(st[j]==')'){if(a[i].l<=0)a[i].r++;  //没有紧邻的（来配对，（ 数量加一else{a[i].l--;   //有可以配对的（，消耗一个）。sum++;      //配对成功的组数加一}}elsea[i].l++;}}sort(a,a+n,cmp);int l=0;    //l表示剩余的“（”数for(int i=0;i<n;i++){if(a[i].r<=l){sum+=a[i].r;l-=a[i].r;    //l减少了a[i].r个，用于配对了}else{sum+=l;l=0;         //l都用了配对了}l+=a[i].l;       //a[i]组字符串右边全是（}printf("%d\n",sum*2);}return 0;
}