codeforces_612C.Replace To Make Regular Bracket Sequence（stack）

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本文介绍了一个算法问题，即如何通过最少次数的替换使一个由多种括号组成的字符串成为规范的括号序列。提供了完整的代码实现，并解释了核心逻辑。

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最简单的stack题目，注意最后栈是否为空就好了。#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <stack>
using namespace std;

stack<char>s;
char a[1000010];
int main()
{
    scanf("%s",&a);
    int i,ans=0,flag=1;
    char c;
    for(i=0; a[i]!='\0'; i++)
    {
        if(a[i]=='('||a[i]=='{'||a[i]=='['||a[i]=='<')
        {
            s.push(a[i]);
        }
        if(!s.empty())
        {
            c=s.top();
        }
        else
        {
            flag=0;
            break;
        }
        if(a[i]==')')
        {
           // printf("%c %c\n",c,a[i]);
            if(c=='{'||c=='['||c=='<')
            {
                ans++;
            }
            else if(c!='(')
            {
                flag=0;
                break;
            }
            s.pop();
        }
        else if(a[i]==']')
        {
           // printf("%c %c\n",c,a[i]);
            if(c=='{'||c=='('||c=='<')
            {
                ans++;
            }
            else if(c!='[')
            {
                flag=0;
                break;
            }
            s.pop();
        }
        else if(a[i]=='}')
        {
           // printf("%c %c\n",c,a[i]);
            if(c=='['||c=='('||c=='<')
            {
                ans++;
            }
            else if(c!='{')
            {
                flag=0;
                break;
            }
            s.pop();
        }
        else if(a[i]=='>')
        {
           // printf("%c %c\n",c,a[i]);
            if(c=='['||c=='('||c=='{')
            {
                ans++;
            }
            else if(c!='<')
            {
                flag=0;
                break;
            }
            s.pop();
        }
    }
    if(!s.empty())
    {
        flag=0;
    }
    if(flag)
        printf("%d\n",ans);
    else
        printf("Impossible\n");
    return 0;
}