【CodeForces】612C - Replace To Make Regular Bracket Sequence（栈，括号配对问题）

C. Replace To Make Regular Bracket Sequence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

[<}){}

output

2

input

{()}[]

output

0

input

]]

output

Impossible

栈括号配对的题，本来不难，刚开始还是WA了。后来添加了两个变量 left 和 right ，记录左右括号的数量，若不相等，则说明没有完成配对。

代码如下：

#include <cstdio>
#include <queue>
#include <stack>
#include <algorithm>
#include <cstring>
using namespace std;
char a[1000011];
int main()
{
	int l;
	int ans;
	bool dot;
	int left,right;
	while (~scanf ("%s",a))
	{
		l=strlen(a);
		stack<char>s;
		dot=true;
		ans=0;
		int left=0,right=0;
		for (int i=0;i<l;i++)
		{
			if (a[i]=='(' || a[i]=='<' || a[i]=='{' || a[i]=='[')
			{
				s.push(a[i]);
				left++;
			}
			else if (a[i]==')')
			{
				right++;
				if (s.empty())
				{
					dot=false;
					break;
				}
				else if (s.top()=='(')
				{
					s.pop();
				}
				else
				{
					ans++;
					s.pop();
				}
			}
			else if (a[i]==']')
			{
				right++;
				if (s.empty())
				{
					dot=false;
					break;
				}
				else if (s.top()=='[')
				{
					s.pop();
				}
				else
				{
					ans++;
					s.pop();
				}
			}
			else if (a[i]=='}')
			{
				right++;
				if (s.empty())
				{
					dot=false;
					break;
				}
				else if (s.top()=='{')
				{
					s.pop();
				}
				else
				{
					ans++;
					s.pop();
				}
			}
			else if (a[i]=='>')
			{
				right++;
				if (s.empty())
				{
					dot=false;
					break;
				}
				else if (s.top()=='<')
				{
					s.pop();
				}
				else
				{
					ans++;
					s.pop();
				}
			}
		}
		if (left!=right)
			dot=false;
		if (dot)
			printf ("%d\n",ans);
		else
			printf ("Impossible\n");
	}
	return 0;
}