HDU 5479 Scaena Felix【STL】-优快云博客

本文介绍了一道算法题目，旨在求解给定括号序列至少需要翻转多少次才能确保任意非空子串都不形成正确匹配的括号对。通过使用标准模板库(STL)中的栈来实现解决方案。

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Scaena Felix
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1070 Accepted Submission(s): 466

Problem Description

Given a parentheses sequence consist of ‘(’ and ‘)’, a modify can filp a parentheses, changing ‘(’ to ‘)’ or ‘)’ to ‘(‘.

If we want every not empty substring of this parentheses sequence not to be “paren-matching”, how many times at least to modify this parentheses sequence?

For example, “()”,”(())”,”()()” are “paren-matching” strings, but “((“, “)(“, “((()” are not.

Input

The first line of the input is a integer T, meaning that there are T test cases.

Every test cases contains a parentheses sequence S only consists of ‘(’ and ‘)’.

1≤|S|≤1,000.

Output

For every test case output the least number of modification.

Sample Input

3
()
((((
(())

Sample Output

1
0
2

Source

BestCoder Round #57 (div.2)

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题意：求正确配对的数目。
解：STL栈。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<stack>
using namespace std;
char str[1010];
int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        int ans = 0;
        scanf("%s", str);
        stack<int>S;
        for (int i = 0; i < strlen(str); i++)
        {
            if (str[i] == '(')
            {
                S.push(1);
            }
            else
            {
                if (!S.empty())
                {
                    ans++;
                    S.pop();
                }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}