2017年上海金马五校程序设计竞赛（网上资格赛）Problem A : Corn's new language

Problem A : Corn's new language

From: DHUOJ, 2017052401

Time Limit: 1 s

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Description

Corn is going to promote programming in the campus, so he wants to add a lot of interesting ideas to make programming more attractive. One task he is working on is to develop a new programming language because he thinks all existing ones are too simple to him. The syntax rules of the language are:

 

1. ()() is a valid program

2. (P)(P) is a valid program, if PP is a valid program

3. PQPQ is a valid program, if both PP and QQ are valid programs

 

Corn wants a compiler for the language. For a program, if it is valid, the compiler should print the max depth in the program. For example "(())(())" has a depth of 22, while "((())())((())())" has a depth of 33. Formally, the max depth is the max number of unmatched open brackets in any prefix.

 

Input

Each case is a non-empty string in a single line, the string will only contain '((' or '))', its length will be no more than 100100.

 

Output

For each case, output "YES" and the integer required above if the program is valid, separated with a space. Or "NO" if the program is invalid.

 

Sample Input

()()(())((())())())((

 

Sample Output

NONOYES 1YES 2YES 3NONO

分析

这道题是在括号匹配的基础上加上了一个深度的记录，这个深度记录就可以简单地定义一个变量初始为0记录深度，再定义一个变量记录最大深度，然后从头到尾来判断这个字符串中括号的方向，如果为‘（’则该变量++，如果为‘）’则该变量--，每次改变后与最大深度比较。

代码

#include <algorithm>#include <iostream>#include <string>#include <vector>#include <stack>#include <queue>#include <set>#include <map>#include <cstdlib>#include <cstring>#include <cstdio>#include <cmath>using namespace std;int main(){    char str[110];    while(scanf("%s",str)!=EOF)    {        int maxn=0,num=0;//maxn存放最大深度        for(int i=0;i<strlen(str);i++)        {            if(str[i]=='(')num++;            if(str[i]==')')num--;            maxn=max(maxn,num);//将各个位置的深度与最大深度对比，然后存放大的那个值        }        stack<char>s;        int len=strlen(str);        for(int i=0;i<len;i++)        {            if(s.empty())                {                    s.push(str[i]);                }            else            {                if(s.top()+1==str[i]||s.top()+2==str[i])//判断栈内元素是否与下一个即将入栈的元素匹配                {                    s.pop();//如果匹配将栈顶元素出栈                }                else                {                    s.push(str[i]);//如果不匹配，则下一个元素入栈                }            }        }        if(s.empty())//如果栈为空，说明括号都已经匹配        {            printf("YES %d\n",maxn);        }        else        {            printf("NO\n");        }    }    return 0;}