[洛谷]P3056 [USACO12NOV]笨牛Clumsy Cows (#贪心 1.17)

题目描述

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:

)( ())( ((())))

给出一个偶数长度的括号序列，问最少修改多少个括号可以使其平衡。

输入输出格式

输入格式：

* Line 1: A string of parentheses of even length at most 100,000 characters.

输出格式：

* Line 1: A single integer giving the minimum number of parentheses that must be toggled to convert the string into a balanced string.

输入输出样例

输入样例#1

())( 

输出样例#1

2 

说明

The last parenthesis must be toggled, and so must one of the two middle right parentheses.

---

思路

#include <stdio.h>
#include <iostream>
using namespace std;
int len,s;
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int i,j,s(0),now(0);
	char c;
	while(cin>>c)//逐个读入字符，这就相当于贪心的子问题。 
	{
		if(c=='(')//使用用个临时计数器，如果是(，就+1，相当于削了一个） 
		{
			now++;
		}
		else if(c==')' && now>0)//削( 
		{
			now--;
		}
		else//如果计算器等于0，就相当于括号全匹配了 。但是当前输入的c怎么处理？就改变括号。 
		{
			s++;//算符+1 
			now++;
		}
	}
	s=s+(now+1)/2;//有now个(没有匹配，将其中的一般转换成）。 
	cout<<s<<endl;
	return 0;
}