Replace To Make Regular Bracket Sequence(括号配对问题)

Replace To Make Regular Bracket Sequence

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u
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Description

You are given string s consists of opening and closing brackets of four kinds <>{}[](). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can’t replace it by ) or >.

The following definition of a regular bracket sequence is well-known, so you can be familiar with it.

Let’s define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2{s1}s2[s1]s2,(s1)s2 are also RBS.

For example the string “[[(){}]<>]” is RBS, but the strings “[)()” and “][()()” are not.

Determine the least number of replaces to make the string s RBS.

Input

The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.

Output

If it’s impossible to get RBS from s print Impossible.

Otherwise print the least number of replaces needed to get RBS from s.

Sample Input

Input

[<}){}

Output

2

Input

{()}[]

Output

0

Input

]]

Output

Impossible
算法的设计思想:

1)凡出现左括弧,则进栈;

2)凡出现右括弧,首先检查栈是否空
   若栈空,则表明该“右括弧”多余,
   否则
栈顶元素
比较,
       若相匹配,则“左括弧出栈” ,
       否则表明不匹配。
3)表达式检验结束时,
   若栈空,则表明表达式中匹配正确,
   否则表明“左括弧”有余
代码:

#include<stdio.h>
#include<string.h>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
char a[1000000+11];
int main()
{
	int len;
	while(scanf("%s",a)!=EOF)
	{
		stack<char>s;
		int left=0,right=0,ans=0;
		bool flag=true;
		len=strlen(a);
		for(int i=0;i<len;i++)
		{
			if(a[i]=='('||a[i]=='{'||a[i]=='['||a[i]=='<')
			{
				s.push(a[i]);
				left++;
			}
			else if(a[i]==')')
			{
				right++;
				if(s.empty())
				{
					flag=false;
					break;
				}
				else
				{
					if(s.top()=='(')
					   s.pop();
					else
					{
						ans++;
						s.pop();
					}
				}
			}
			else if(a[i]=='}')
			{
				right++;
				if(s.empty())
				{
					flag=false;
					break;
				}
				else
				{
					if(s.top()=='{')
					   s.pop();
					else
					{
						ans++;
						s.pop();
					}
				}
			}
			else if(a[i]==']')
			{
				right++;
				if(s.empty())
				{
					flag=false;
					break;
				}
				else
				{
					if(s.top()=='[')
					{
						s.pop();
					}
					else
					{
						ans++;
						s.pop();
					}
				}
			}
			else if(a[i]=='>')
			{
				right++;
				if(s.empty())
				{
					flag=false;
					break;
				}
				else
				{
					if(s.top()=='<')
					{
						s.pop();
					}
					else
					{
						ans++;
						s.pop();
					}
				}
			}
		}
		if(right!=left)
		   flag=false;
		if(flag)
		   printf("%d\n",ans);
		else
		   printf("Impossible\n");
	}
	return 0;
}




    原文作者:括号匹配问题
    原文地址: https://blog.csdn.net/qq_34681949/article/details/52025758
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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