CSU - 1271 Brackets Sequence

括号序列修复算法

最新推荐文章于 2020-03-20 21:35:48 发布

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最新推荐文章于 2020-03-20 21:35:48 发布

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本文链接：https://blog.youkuaiyun.com/l18339702017/article/details/79938365

本文介绍了一种算法，用于确定如何通过插入最少数量的括号将不规则的括号序列转换为规则序列。该算法考虑了左括号和右括号的数量差异，并提供了解决方案以最小化所需的插入次数。

Let us define a regular brackets sequence in the following way:

1. Empty sequence is a regular sequence.

2. If S is a regular sequence, then (S) is a regular sequence.

3. If A and B are regular sequences, then AB is a regular sequence.

For example, these sequences of characters are regular brackets sequences: (), (()), ()(), ()(()), ((())())().

And all the following character sequences are not: (, ), ((), ()), ())(, (()(, ()))().

A sequence of characters '(' and ')' is given. You can insert only one '(' or ')' into the left of the sequence, the right of the sequence, or the place between any two adjacent characters, to try changing this sequence to a regular brackets sequence.

Input

The first line has a integer T (1 <= T <= 200), means there are T test cases in total.

For each test case, there is a sequence of characters '(' and ')' in one line. The length of the sequence is in range [1, 10⁵].

Output

For each test case, print how many places there are, into which you insert a '(' or ')', can change the sequence to a regular brackets sequence.

What's more, you can assume there has at least one such place.

Sample Input

4
)
())
(()(())
((())())(()

Sample Output

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#define LL long long
using namespace std;

const int MAX = 1e6+10;
const int INF = 0x3fffffff;

int a[MAX];
char s[MAX];

int main(){
    int t;
    cin>>t;
    getchar();
    while(t--){
        memset(a,0,sizeof(a));
        memset(s,0,sizeof(s));
        gets(s);
        int ln = 0,rn = 0;
        int len = strlen(s);
        for(int i=0;i<len;i++){
            if(s[i]=='('){
                a[i] = -1;
                ln++;
            }
            else{
                rn++;
                a[i] = 1;
            }
        }
        if(rn<ln){//缺右括号 
            int sum = 0;
            int temp =0;
            for(int i=0;i<len;i++){
                sum+=a[i];
                if(sum<0)
                	temp++;
				else if (sum==0)
					temp = 0;
            }
            cout<<temp<<endl;
        }
        else{
            int sum =0;
            int temp = 0;
            for(int i=len-1;i>=0;i--){
            	sum+=a[i];
                if(sum > 0)  
                    temp++;  
                else if(sum == 0)  
                    temp=0; 
            }
            cout<<temp<<endl;
        }
    }
    return 0;
}