hduSeinfeld

Seinfeld

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1405    Accepted Submission(s): 695

Problem Description

I’m out of stories. For years I’ve been writing stories, some rather silly, just to make simple problems look difficult and complex problems look easy. But, alas, not for this one.
You’re given a non empty string made in its entirety from opening and closing braces. Your task is to find the minimum number of “operations” needed to make the string stable. The definition for being stable is as follows:
1. An empty string is stable.
2. If S is stable, then {S} is also stable.
3. If S and T are both stable, then ST (the concatenation of the two) is also stable.
All of these strings are stable: {}, {}{}, and {{}{}}; But none of these: }{, {{}{, nor {}{.
The only operation allowed on the string is to replace an opening brace with a closing brace, or visa-versa.

 

Input

Your program will be tested on one or more data sets. Each data set is described on a single line. The line is a non-empty string of opening and closing braces and nothing else. No string has more than 2000 braces. All sequences are of even length.
The last line of the input is made of one or more ’-’ (minus signs.)

 

Output

For each test case, print the following line:
k. N
Where k is the test case number (starting at one,) and N is the minimum number of operations needed to convert the given string into a balanced one.
Note: There is a blank space before N.

 

Sample Input

  

   }{
{}{}{}
{{{}
---
  

 

Sample Output

  

   1. 2
2. 0
3. 1
  
  


  
  

   题目归结起来就是括号的配对
  
  

   做这道题目有很多种方法
  
  

   我参看别人的代码，
  
  

   而且这是第一次使用栈
  
  

   代码如下：
  
  

   
#include<stdio.h>
#include<stack>
using namespace std;
char str[1200];
int main(){
	int m=0;
	char r,l;
	stack<char>s;     //建立一个栈 
	while(~scanf("%s",str),str[0]!='-'){
		for(int i=0;str[i]!='\0';i++){
			if(str[i]=='}'){
				if(!s.empty()&&s.top()=='{')
				   s.pop();          //empty()是判断是否是空值 
				else                 //pop是消值 
				   s.push(str[i]);    //push是存入栈 
			}
			else  s.push(str[i]);
		}
		int ans=0;
		while(!s.empty()){
			r=s.top();        //取得最前面的字符 
			s.pop();          //然后删除 
			l=s.top();       //取得最后一个的字符 
			s.pop();         //删除 
			if(r=='{')   ans++;
			if(l=='}')   ans++;
		}
		printf("%d. %d\n",++m,ans);
	}
	return 0;
}