HDU 5479(栈的应用)

Scaena Felix

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 426 Accepted Submission(s): 186

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

//判断（）的个数行
// 利用栈的先进后出的特点模拟

#include <stdio.h>
#include <string.h>
#include <stack>
using namespace std;
char str[1000+10];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
getchar();
scanf("%s",str);
stack <char> s;
int len=strlen(str);
int cnt=0;
for(int i=0;i<len;i++)
{
if(str[i]=='(')
s.push(str[i]);
else if(str[i]==')')
{
if(!s.empty())  //判断是否为空
{
s.pop();
cnt++;
}
}
}
printf("%d\n",cnt);
}
return 0;
}