CF5C Longest Regular Bracket Sequence

题目来源CodeForces 5C

评测方式RemoteJudge

难度提高+/省选-

题意翻译

给出一个括号序列，求出最长合法子串和它的数量。 合法的定义：这个序列中左右括号匹配

题目描述

This is yet another problem dealing with regular bracket sequences.

We should remind you that a bracket sequence is called regular, if by inserting «+» and «1» into it we can get a correct mathematical expression. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.

You are given a string of «(» and «)» characters. You are to find its longest substring that is a regular bracket sequence. You are to find the number of such substrings as well.

输入格式

The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed $10^{6}$ .

输出格式

Print the length of the longest substring that is a regular bracket sequence, and the number of such substrings. If there are no such substrings, write the only line containing "0 1".

输入输出样例

输入 #1

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

输出 #1

6 2

输入 #2

))(

输出 #2

0 1

#include <bits/stdc++.h>

using namespace std;

const int M = 1e6 + 5;

char s[M];

int n[M];

stack<int> st;

int main()

{

scanf("%s", s + 1);

int l = strlen(s + 1);

int ml = 0, mn = 1;

for (int i = 1; i <= l; i++)

{

if (st.empty() && s[i] == ')')

continue;

if (s[i] == ')')

{

n[i] = i - st.top() + 1 + n[st.top() - 1];

st.pop();

if (n[i] == ml)

mn++;

if (n[i] > ml)

{

ml = n[i];

mn = 1;

}

}

else

st.push(i);

}

printf("%d %d\n", ml, mn);

return 0;

}