Educational Codeforces Round 4 C. Replace To Make Regular Bracket Sequence

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C. Replace To Make Regular Bracket Sequence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ands₂ 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 exceed 10⁶.

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 test(s)

Input

[<}){}

Output

Input

{()}[]

Output

Input

]]

Output

Impossible

题意：<>, {}, [], ()都是RBS，<RBS>RBS, {RBS}RBS, [RBS]RBS, (RBS)RBS也是RBS，意思就是如果<>, {}, [], ()里包含有字符串那么被包含的字符串也得失RBS。现在有一段字符串，称'<'、'{'、'['、'('为开类型，反之为闭类型，只有同类型之间可以自由转换，现在问最少得转换多少次才使得整个字符串都符合RBS，如过不行就输出“Impossible”.

思路：用栈模拟，遍历字符串数组，只能放进开类型，当遇到闭类型时，如果栈顶的字符是和它匹配的就不用转换，否则需要转换；当遇到闭类型时，如果栈为空的话，这个时候就是“Impossible”，因为闭类型不能转换成开类型。最后如果栈不为空也是“Impossible”。当时理解错题意了。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
#include <stack>
using namespace std;
#define maxn 1000005
char s[maxn];
char mapping(char a)
{
    if(a=='{') return '}';
    if(a=='<') return '>';
    if(a=='[') return ']';
    if(a=='(') return ')';
}
int main()
{
    stack<char> st;
    scanf("%s", s);
    int len = strlen(s);
    int ans = 0;
    for(int i = 0;i < len;i++)
    {
        if(s[i]=='<'||s[i]=='{'||s[i]=='['||s[i]=='(') st.push(s[i]);
        else if(!st.empty()){
            if(mapping(st.top()) != s[i])
                ans++;
            st.pop();
        }else{
            printf("Impossible\n");
            return 0;
        }
    }
    if(!st.empty()) printf("Impossible\n");
    else printf("%d\n", ans);
}