codeforces 5C Longest Regular Bracket Sequence -- 贪心

C. Longest Regular Bracket Sequence

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 106.

Output

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".

Sample test(s)

input

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

output

6 2

input

))(

output

0 1

这一题想到思路就很简单。重点是找），离它最近的没有被匹配的（就是应该跟它匹配，可以记录这一对有多长。然后，（）（），对于这种情况，就是两队相邻，第二队加上第一队长度就好了。

#include<cstdio>#include<iostream>#include<stack>using namespace std;#define MAXN 1000000int len[MAXN]={0};int main(){#ifndef ONLINE_JUDGE  freopen("data","r",stdin);#endif  string s;  stack<int> st;  int k,maxl=0,n=0;  cin>>s;  for(int i=0; i<s.size(); i++){    if('(' == s[i]){      st.push(i);    }else{      if(!st.empty()){        k = st.top();        st.pop();        len[i] = ( k>0 ? len[k-1] : 0 ) + i+1-k;        if(len[i]>maxl){          maxl = len[i];          n = 1;        }else if(len[i]==maxl)          n++;      }    }  }  if(maxl>0)    cout<< maxl << " " << n << endl;  else    cout<<"0 1"<<endl;#ifndef ONLINE_JUDGE  fclose(stdin);#endif  return 0;}

之前一个错误的思路就是从第一字符到后面的字符，分别找最大长度，顺着数（，用）来抵消，这样在类似（）（（）（（）（这种情况下，每次都要走完，复杂度为n平方，时间过不去。而贪心的方法复杂度仅为n。