G - Replace To Make Regular Bracket Sequence

最新推荐文章于 2022-01-18 10:40:24 发布

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最新推荐文章于 2022-01-18 10:40:24 发布

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分类专栏：栈和队列

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本文介绍了一个算法问题，即如何通过最少次数的替换操作使给定的由四种类型的括号组成的字符串变成一个有效的常规括号序列（RBS）。文章提供了一个C++实现方案，详细展示了如何遍历输入字符串并使用栈来跟踪括号匹配的过程。

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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.

Example

Input

[<}){}

Output

Input

{()}[]

Output

Input

]]

Output

Impossible

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
char a[1000002];
using namespace std;
int main()
{
	while(~scanf("%s",a))
	{
	 stack<char > s;
	 int i,t=0,n=strlen(a),k=0;
	 for(i=0;i<n;i++)
	{
	  if(a[i]=='{'||a[i]=='['||a[i]=='<'||a[i]=='(')    //将左括号压入栈内
	    s.push(a[i]);
      else
	  {
      	if(s.empty())  //左括号没了，但是还有右括号
      	{
      	  k=1;
      	  break;
        }
        else
        {
		 if(a[i]-s.top()==2||a[i]-s.top()==1)    //若是一对括号，差值为1或2
		   s.pop();
		 else
		 {
           t++;
	       s.pop();
	     }
	    }
	  }
    }
    if(k==1||!s.empty())
      printf("Impossible\n");
    else
      printf("%d\n",t);
	}
	return 0;
}