HDU5479 Scaena Felix【堆栈+输入流】

Scaena Felix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1041    Accepted Submission(s): 450

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 <b>substring</b> 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)

问题链接：HDU5479 Scaena Felix

问题简述：参见上文。

问题分析：

这个题就是求输入字符串中有多少括号是匹配的。

这类匹配问题看似堆栈问题，是需要堆栈的概念的，即遇到左括号则进栈遇到右括号与栈中的括号匹配一下。

然而，实际上没有必要用堆栈，计数一下就好了。

程序说明：按字符流读入数据进行处理是最佳选择，使用缓存那是浮云。

题记：（略）

参考链接：（略）

AC的C语言程序如下：

/* HDU5479 Scaena Felix */

#include <stdio.h>

int main(void)
{
    int t;
    char c;

    scanf("%d", &t);
    getchar();
    while(t--) {
        int left = 0, ans = 0;

        while((c = getchar()) != '\n') {
            if(c == '(')
                left++;
            else if(c == ')') {
                if(left > 0) {
                    left--;
                    ans++;
                }
            }
        }

        printf("%d\n", ans);
    }

    return 0;
}