ANARC09A - Seinfeld

I’m out of stories. For years I’ve been writing stories, some rather silly, just to make simple problems look difficult and complex problems look easy. But, alas, not for this one.

You’re given a non empty string made in its entirety from opening and closing braces. Your task is to find the minimum number of “operations” needed to make the string stable. The definition for being stable is as follows:

  1. An empty string is stable.
  2. If S is stable, then {S} is also stable.
  3. If S and T are both stable, then ST (the concatenation of the two) is also stable.

All of these strings are stable: {}, {}{}, and {{}{}}; But none of these: }{, {{}{, nor {}{.
The only operation allowed on the string is to replace an opening brace with a closing brace, or vice-versa.

Input

Your program will be tested on one or more data sets. Each data set is described on a single line. The line is a non-empty string of opening and closing braces and nothing else. No string has more than 2000 braces. All sequences are of even length.

The last line of the input is made of one or more ’-’ (minus signs.)

Output

For each test case, print the following line:

k. N

Where k is the test case number (starting at one,) and N is the minimum number of operations needed to convert the given string into a balanced one.

Example

Input:
}{
{}{}{}
{{{}
---

Output: 1. 2
2. 0
3. 1

Added by:Mohammad Kotb
Date:2009-11-28
Time limit:3.236s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64 BASH JS-RHINO
Resource:http://www.icpc-anarc.org

hide comments
2018-09-06 18:21:21
using stack will also give you correct ans becoz
for every stable string it will be form of SS e.g. {}{} or {S} e.g. {{}}
but in both cases no of { will be equal to }. So to make a string stable either you make you it some form of {}{} or {{}} both will give you same ans. in both cases you will be doing only inversion of braces. Just don't invert already stable sub-string.
2018-07-29 07:50:10


Last edit: 2018-07-29 07:56:32
2018-07-01 12:53:35
Why is it tagged DP?????
2018-06-17 15:59:20
Use getline to read input. Adhoc solution is possible. But those who want to know how to solve using dp the states are dp(n,unbalance), where n is my current index and unbalance is unbalance upto current index. One can easily prove that adhoc solution is correct indeed.
2018-06-11 17:12:12
AC in 1 go is easily possible with a stack. Even without a stack, simply taking the count of brackets can easily solve this problem. Very trivial problem. No need to waste time in getting a DP solution working :)
2018-06-07 13:40:45
Can anyone tell me how to solve it using dp??
2018-06-06 12:27:32
@puru it's given that length of the string is even
2018-06-04 20:14:16
what should be the output of {}{
2018-05-17 14:17:45
can anyone provide me a test case where there is a possibility of wrong answer??
2018-05-13 16:37:47
Good question on stack


Last edit: 2018-05-14 09:19:44
© Spoj.com. All Rights Reserved. Spoj uses Sphere Engine™ © by Sphere Research Labs.