hdoj3351-stack

Problem Description

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 visa-versa.

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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.)

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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.
Note: There is a blank space before N.

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Sample Input

}{ {}{}{} {{{} ---

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Sample Output

1. 2 2. 0 3. 1

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思路：

一開始設置了兩個變量，l和r分別代表左括號和右括號，后來發現和stack相關的操作只用設置一個變量即可，可是說是用來記錄stack高度的也可以說是用來記錄'{'值的。這樣不僅更方便靈活，而且（我想不到合適的詞語來描述。。）更容易激發靈感！

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#include <iostream>
#include <string>
using namespace std;int main()
{string ss;int shift,l,K = 0;while(cin>>ss) {if(ss[0] == '-')break;l = 0;shift = 0;for(int i = 0;i < ss.length();i++) {if(ss[i] == '{') l++;else if(l) l--;else {l++;shift++;}}cout<<++K<<". "<<l/2+shift<<endl;}return 0;
}

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