zoukankan      html  css  js  c++  java
  • Seinfeld HDU

    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. 

    InputYour 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.) 

    OutputFor 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. 
    Sample Input

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

    Sample Output 1. 2

    2. 0
    3. 1
    把 满足条件的全都扔出来
    就只存在 这三种情况
     1 #include <iostream>
     2 using namespace std;
     3 #include<string.h>
     4 #include<string>
     5 #include<set>
     6 #include<sstream>
     7 #include<stdio.h>
     8 #include<queue>
     9 #include<math.h>
    10 #include<algorithm>
    11 #include<stack>
    12 int main()
    13 {
    14     stack<char>s;
    15     char a[2010],j,b[2010];
    16     int i;
    17     int add=0;
    18     while(gets(a))
    19     {
    20 
    21         if(a[0]=='-'&&a[1]=='-'&&a[2]=='-')
    22             break;
    23         cout<<++add<<". ";
    24 
    25         memset(b,0,sizeof(b));
    26         int lena=strlen(a);
    27 
    28         for(i=0;i<lena;i++)
    29         {
    30             if(a[i]=='{')
    31                 s.push('{');
    32             else if(s.empty()&&a[i]=='}')
    33                 s.push('}');
    34             else if(!s.empty()&&s.top()=='}'&&a[i]=='}')
    35                 s.push('}');
    36             else if(!s.empty()&&s.top()=='{'&&a[i]=='}')
    37                 s.pop();
    38         }
    39         i=0;
    40         while(!s.empty())
    41         {
    42             b[i++]=s.top();
    43             s.pop();
    44         }
    45         int lenb=strlen(b);
    46         int sum=0;
    47         for(i=0;i<lenb;i=i+2)
    48         {
    49             if(b[i]==b[i+1])
    50                 sum++;
    51             else
    52                 sum=sum+2;
    53         }
    54         cout<<sum<<endl;
    55     }
    56     return 0;
    57 }
    View Code

    }}}}}}}}}}}}}
    {{{{{{{{{{{
    }}}}}}}}{{{{{{{{{{
  • 相关阅读:
    刚刚开通
    腾讯面试经历2015
    排序之归并排序
    AC自动机
    后缀数组初步
    概率dp初探
    【NOIP2015】反思+题解
    Built-in functions
    poj2528 Mayor's posters(线段树区间覆盖)
    Codeforces #317 C.Lengthening Sticks(数学)
  • 原文地址:https://www.cnblogs.com/dulute/p/7272617.html
Copyright © 2011-2022 走看看