Codeforces 612C Replace To Make Regular Bracket Sequence【栈】

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 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

这是栈的括号配对问题，就像南阳OJ的题型，但是不同的是这道题中，不同类型的括号还可以互换，而且要是可以全部出栈，输出替换的次数。

#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAX_N 1000005
using namespace std;
const int INF = 0xffff;
char s[MAX_N];

//用数组将每种括号存储下来，对于判断括号的类型就可以借助数组下标，比较方便
char c[] = {'<', '>', '(', ')', '{', '}', '[', ']'};
int main() {
    stack<char> stk;
    while (scanf("%s", s) != EOF) {
        while (!stk.empty())  stk.pop();
        int ans = 0;
        int len = strlen(s), k, m, n;
        for (int i = 0; i < len; i++) {
            if (!stk.empty()) {
            	
				//查找字符在数组中的下标，确定字符的类型
                for (k = 0; k < 8; k++) {
                    if (stk.top() == c[k])  m = k;
                    if (s[i] == c[k])   n = k;

                }
                //类型不同并且出栈元素为左边的括号才可以出栈
                if (m%2 != n%2 && !(m%2)) {
                    if (m != n - 1) ans++;
                    stk.pop();
                }
                else stk.push(s[i]);
            }
            else stk.push(s[i]);
        }
        if (stk.empty())  printf("%d
", ans);
        else printf("Impossible
");
    }
    return 0;
}