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

Replace To Make Regular Bracket SequenceCodeForces - 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

2

Input

{()}[]

Output

0

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);elseprintf("Impossible\n");}return 0;}