Codeforces 612C： Replace To Make Regular Bracket Sequence（栈）

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

题目大意：括号匹配的变形题，在这里只要是左括号，就能跟右括号消掉，但是消掉前得把左括号变成和右括号一样的才能消掉，问给你一个字符串，匹配的时候需要将左括号变几次，如果不能消掉，则输出Impossible。
解题思路：跟括号匹配差不多，就是多了几个条件。
代码如下：

#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
using namespace std;
char a[1000010];
int main()
{
scanf("%s",a);
int size=strlen(a);
stack<char>q;
int cnt=0;
for(int i=0;i<size;i++)
{
if(q.empty())
{
q.push(a[i]);
}
else
{
if((q.top()=='<'&&(a[i]=='>'||a[i]==')'||a[i]=='}'||a[i]==']'))||(q.top()=='('&&(a[i]=='>'||a[i]==')'||a[i]=='}'||a[i]==']'))||(q.top()=='{'&&(a[i]=='>'||a[i]==')'||a[i]=='}'||a[i]==']'))||(q.top()=='['&&(a[i]=='>'||a[i]==')'||a[i]=='}'||a[i]==']')))//左括号碰到了右括号
{
if((q.top()=='<'&&a[i]=='>')||(q.top()=='('&&a[i]==')')||(q.top()=='{'&&a[i]=='}')||(q.top()=='['&&a[i]==']'))//如果左右括号是同类的，那么直接消掉
{
q.pop();
}
else
{
cnt++;//不同类的，消掉前，变数+1
q.pop();
}
}
else//左右括号不同类，加入栈
{
q.push(a[i]);
}
}
}
if(q.empty())//如果最后匹配消掉成功，则栈一定是空的
printf("%d\n",cnt);
else
printf("Impossible\n");
return 0;
}