【CodeForces】612C---Replace To Make Regular Bracket Sequence（栈）

最新推荐文章于 2022-03-17 15:16:34 发布

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最新推荐文章于 2022-03-17 15:16:34 发布

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分类专栏：栈与队列

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这篇博客介绍了如何通过替换操作使非正规括号序列变为正规，重点聚焦于利用栈的数据结构来解决CodeForces上的612C题目。

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Replace To Make Regular Bracket Sequence CodeForces - 612C

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

思路分析：括号配对问题加强版，关键判断最后栈是否为空。

代码如下：

#include<cstdio>
#include<cstring>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
char str[1000000+11];
int main()
{
	stack<char> s;			//栈 
	while(scanf("%s",str)!=EOF)
	{
		bool ans=true;
		int num=0;
		int l=strlen(str);
		for(int i=0;i<l;i++)
		{
			if(str[i]=='{'||str[i]=='['||str[i]=='('||str[i]=='<')		//是左括号直接入栈 
				s.push(str[i]);
			else if(str[i]=='}')
			{
				if(s.empty() )//此时栈为空，让'}'进栈，那么栈就不空了，但是如此以后栈再也不会变空了，所以直接结束循环了可以 
				{
					s.push(str[i]);
					break; 
				}
				else if(s.top() =='{')//直接就可以配对成功，'{'直接弹出，'}'并不用进栈 
					s.pop() ;
				else if(s.top() =='['||s.top() =='('||s.top() =='<')//说明左右括号可以通过转换配对成功 
				{
					s.pop() ;				//直接弹出栈顶元素，'}'并不进栈 
					num++;					//统计转换的次数 
				}
				else					//说明是其他情况，根本不符合题意，可以直接结束循环 
				{
					ans=false;
					break;
				}
			} 
			else if(str[i]==']')			//下面的理由同上 
			{
				if(s.empty() )
				{
					s.push(str[i]);
					break; 
				}
				else if(s.top() =='[')
					s.pop() ;
				else if(s.top() =='{'||s.top() =='('||s.top() =='<')
				{
					s.pop() ;
					num++;
				}
				else
				{
					ans=false;
					break;
				}
			}
			else if(str[i]==')')
			{
				if(s.empty() )
				{
					s.push(str[i]);
					break; 
				}
				else if(s.top() =='(')
					s.pop() ;
				else if(s.top() =='['||s.top() =='{'||s.top() =='<')
				{
					s.pop() ;
					num++;
				}
				else
				{
					ans=false;
					break;
				}
			}
			else if(str[i]=='>')
			{
				if(s.empty() )
				{
					s.push(str[i]);
					break; 
				}
				else if(s.top() =='<')
					s.pop() ;
				else if(s.top() =='['||s.top() =='('||s.top() =='{')
				{
					s.pop() ;
					num++;
				}
				else
				{
					ans=false;
					break;
				}
			}
		}
		if(!s.empty() )
			ans=false;
		if(ans)				//关键判断：最后栈为空才返回值 ，不空说明不能配对成功 
			printf("%d\n",num);
		else
			printf("Impossible\n");
	}
	return 0;
}