Codeforces-612C-Replace To Make Regular Bracket Sequence 【stack】

C. Replace To Make Regular Bracket Sequence

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

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.

Sample Input

Input

[<}){}

Output

2

Input

{()}[]

Output

0

Input

]]

Output

Impossible

题意：给定一个只有'{' '[' '(' '<'开字符和'}' ']' ')' '>'闭字符的字符串，你可以将开字符互相转化，闭字符互相转化，要求得到一个括号匹配的串。若可以输出最少转化次数，反之输出Impossible。

思路：就是栈而已

#include<cstdio>#include<stack>#include<string.h>#define M 1000010using namespace std;char a[M];int main(){int len,ans;while(scanf("%s",&a)!=EOF){ans=0;stack<char> st;len=strlen(a);for(int i=0;i<len;i++){if(a[i]=='>'||a[i]==']'||a[i]=='}'){if(st.empty()){ printf("Impossible\n"); ans=-1; break;}if(st.top()!=a[i]-2)ans++;st.pop();}else if(a[i] == ')'){if(st.empty()){ printf("Impossible\n"); ans=-1; break;}if(st.top()!='(')ans++;st.pop();}else st.push(a[i]);}if(ans!=-1){if(!st.empty())  printf("Impossible\n"); else printf("%d\n",ans);}}}