DHUOJ 2017052401 - Corn's new language（栈括号匹配）

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最新推荐文章于 2023-04-08 18:06:33 发布

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本文介绍了一种使用栈解决括号匹配问题的方法，并通过一个示例程序详细展示了如何计算括号字符串的最大深度。适用于编程竞赛及面试准备。

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来自：2017年上海金马五校程序设计竞赛（网上资格赛）

Corn's new language

From: DHUOJ, 2017052401

&lt;a type="button" class="btn btn-warning" href="/solution/submit.html?problemId=5261"&gt;Submit&lt;/a&gt;

Time Limit: 1 s

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) is a valid program, if P is a valid program

3. PQ is a valid program, if both P and Q 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 2 , while " ((())()) " has a depth of 3 . 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 100 .

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

NO
NO
YES 1
YES 2
YES 3
NO
NO

Author: Tianyi Chen

思路：

简单的括号匹配问题，栈解决

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<stack>
using namespace std;
 
int main()
{
    char s[1000000];
    while(cin>>s)
    {   
        stack<char>st;
        bool flag=1;
        int maxn=0;
        int len=strlen(s);
        for(int i=0;i<len;i++){
            if(s[i]=='('){
                st.push(s[i]);
                if(maxn<st.size()){
                    maxn=st.size();
                }
            }
            if(s[i]==')'){
                if(st.size()<=0){
                    flag=0;
                    break;
                }
                st.pop();
            }
        }
        if(st.size()>0)cout<<"NO"<<endl;
        else if(flag==0)cout<<"NO"<<endl;
        else{
            cout<<"YES "<<maxn<<endl;
        }
    }
    return 0;   
}