BZOJ3016: [Usaco2012 Nov]Clumsy Cows

3016: [Usaco2012 Nov]Clumsy Cows

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 71  Solved: 52
[Submit][Status]

Description

Bessie the cow is trying to type a balanced string of parentheses into her new laptop, but she is sufficiently clumsy (due to her large hooves) that she keeps mis-typing characters. Please help her by computing the minimum number of characters in the string that one must reverse (e.g., changing a left parenthesis to a right parenthesis, or vice versa) so that the string would become balanced. There are several ways to define what it means for a string of parentheses to be "balanced". Perhaps the simplest definition is that there must be the same total number of ('s and )'s, and for any prefix of the string, there must be at least as many ('s as )'s. For example, the following strings are all balanced:

()

(())

()(()())

while these are not:

)(

())(

((())))

 

问题描述

给定长度为n的一个括号序列，每次修改可以修改一个位置的括号，若这个括号为’(‘，则修改为’)’，若这个括号为’)’，则修改为’(‘，问最少修改多少个使得原括号序列合法。

其中：

①     ()是合法的；

②     若A是合法的，则(A)是合法的；

③     若A，B都是合法的，则AB是合法的。

 

Input

       一个长度为n个括号序列。

 

Output

 

       最少的修改次数。

 

Sample Input

())(

Sample Output

2

样例说明
修改为()()，其中红色部分表示修改的括号。

数据范围
100%的数据满足：1 <= n <= 100,000。

HINT

 

Source

Silver

题解：

不错的贪心。

用s记录前面多出来的'('的个数，每次读入'('s++,读入')'s--

s<0时表示没有‘(’还多出')'就需要把')'变成'('，这样s+=2，ans++

最后如果s!=0的时候还要把一半的'('变成')'，就是ans+=s/2

想的时候因为思路比较乱，不敢大胆想，所以没想出来T_T

代码：

#include<cstdio>
int n,s,ans;
char ch;
int main()
{
    while (scanf("%c",&ch)!=EOF)
      {
          if (ch=='\n')break;
          if (ch=='(')s++;
          if (ch==')')
          {
              s--;
              if (s<0){s+=2;ans++;}
          }
      }
    ans+=s/2;
    printf("%d",ans);
}