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最新推荐文章于 2019-01-28 16:13:00 发布

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本文介绍了一个算法问题，涉及四种类型的括号（圆括号、方括号、尖括号及花括号）。任务是确定最少的操作次数以使输入的括号序列变为有效的正则括号序列，如果无法实现则输出'Impossible'。

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 ofs does not exceed10⁶.

Output

If it's impossible to get RBS from s printImpossible.

Otherwise print the least number of replaces needed to get RBS from s.

Sample Input

Input

[<}){}

Output

Input

{()}[]

Output

Input

]]

Output

Impossible

就是一串字符串，只包含四种括号()，{}, [],<>,左括号可以互相转化，<,{,[,(，右括号可以互相转化，>,},],)，但左右不能转化，最后问所有括号配对最少需要多少步，不能就输出impossible
ps：括号是就近匹配，否则就是这种情况[( ])，这样配对是不行的

#include<cstdio>
#include<cstring>
#include<stack>
#include<queue>
#include<algorithm>
using namespace std;
char s[1000010];
char ope(char a){
    if(a=='{'){
    	return '}';
	}
    if(a=='<'){
    	return '>';
	}
    if(a=='['){
    	return ']';
	}
    if(a=='('){
    	return ')';
	} 
}
int main(){
    stack <char> sta;
    scanf(" %s", s);
    int n = strlen(s);
    int ans = 0;
    for(int i = 0;i < n;i++){//左括号就进栈
        if(s[i] == '<' || s[i] == '{'||s[i] == '[' || s[i] == '('){
        	sta.push(s[i]);
		} 
        else if(!sta.empty()){//如果是右括号，且栈非空，
            if(ope(sta.top()) != s[i]){//就近匹配，对于栈，最近的就是最后进入的，也是栈顶元素
            	ans++;
			}
            sta.pop();
        }
		else{//空栈即右括号多余，就是不能匹配
            printf("Impossible\n");
            return 0;
        }
    }
    if(!sta.empty()){//有多余的左括号，就是不能匹配
    	printf("Impossible\n");
	} 
    else{
    	printf("%d\n", ans);
	}
	return 0;
}