Replace To Make Regular Bracket Sequence 【栈】

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 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

水题，可是我还是wa 1次，忘记考虑 左括号要是多 也是Impossible

代码

#include<bits/stdc++.h>using namespace std;#define LL long long const int MAXN =1e6+10;char s[MAXN];int main(){    gets(s);    stack<char>S; while(!S.empty())  S.pop();    int cnt=0;int flag=1;    for(int i=0;s[i];i++){        if(s[i]=='<'||s[i]=='{'||s[i]=='['||s[i]=='(') S.push(s[i]);        else if(s[i]=='>'){            if(S.empty()) { flag=0;break;}            if(S.top()=='<') {S.pop();}            else { cnt++; S.pop();}         }else if(s[i]=='}'){            if(S.empty()) { flag=0;break;}            if(S.top()=='{') {S.pop();}            else { cnt++; S.pop();}        }else if(s[i]==']') {            if(S.empty()) { flag=0;break;}            if(S.top()=='[') {S.pop();}            else { cnt++; S.pop();}        }else if(s[i]==')'){            if(S.empty()) { flag=0;break;}            if(S.top()=='(') {S.pop();}            else { cnt++; S.pop();}        }    }    if(!flag||!S.empty()) puts("Impossible");    else printf("%d\n",cnt);    return 0;}