Move Brackets CodeForces - 1374C rat1000

给定一个长度为偶数n的括号序列s,由n²个左括号'('和n²个右括号')'组成。任务是找到使s变为正规括号序列所需的最小移动次数。正规括号序列可以通过连接、插入和删除操作获得。题目提供多个测试案例,每个案例给出n和序列s,需要求出最小移动次数。

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You are given a bracket sequence s of length n, where n is even (divisible by two). The string s consists of n2 opening brackets ‘(’ and n2 closing brackets ‘)’.

In one move, you can choose exactly one bracket and move it to the beginning of the string or to the end of the string (i.e. you choose some index i, remove the i-th character of s and insert it before or after all remaining characters of s).

Your task is to find the minimum number of moves required to obtain regular bracket sequence from s. It can be proved that the answer always exists under the given constraints.

Recall what the regular bracket sequence is:

“()” is regular bracket sequence;
if s is regular bracket sequence then “(” + s + “)” is regular bracket sequence;
if s and t are regular bracket sequences then s + t is regular bracket sequence.
For example, “()()”, “(())()”, “(())” and “()” are regular bracket sequences, but “)(”, “()(” and “)))” are not.

You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1≤t≤2000) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (2≤n≤50) — the length of s. It is guaranteed that n is even. The second line of the test case containg the string s consisting of n2 opening and n2 closing brackets.

Output
For each test case, print the answer — the minimum number of moves required to obtain regular bracket sequence from s. It can be proved that the answer always exists under the given constraints.

Example
Input
4
2
)(
4
()()
8
())()()(
10
)))((((())
Output
1
0
1
3
Note
In the first test case of the example, it is sufficient to move the first bracket to the end of the string.

In the third test case of the example, it is sufficient to move the last bracket to the beginning of the string.

In the fourth test case of the example, we can choose last three openning brackets, move them to the beginning of the string and obtain “((()))(())”.
左右配对

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<cstdio>
#include<vector>
#include<set>
#include<cstring>
#include<cstdlib>
#include<time.h>
#include<stack>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
char a[60];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        cin>>n>>a;
        int ans=0,sum=0;
        for(int i=0;i<n;i++){
            if(a[i]=='(')
                ans++;
            else{
                if(ans>0)
                    ans--;
                else
                    sum++;
            }
        }
        cout<<sum<<endl;
    }
  system("pause");
  return 0;
}
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