CodeForces 5C Longest Regular Backet sequence
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".
Examples
)((())))(()())
6 2
))(
0 1
题解:DP。用栈存每一个‘(’的下标,遇到一个')',判断一下栈是否为空,若为空,则继续往后,若不为空,则新增加长度为:i-temp+1;
参考代码为:
#include<bits/stdc++.h> using namespace std; const int maxn=1e6+10; char s[maxn]; int dp[maxn]; int main() { scanf("%s",s+1); stack<int> st; int len=strlen(s+1),Max=1,ans=0; for(int i=1;i<=len;i++) { if(s[i]=='(') st.push(i); else { if(!st.empty()) { int temp=st.top(); st.pop(); dp[i]=dp[temp-1]+i-temp+1; Max=max(dp[i],Max); } } } if(Max==1) printf("0 1\n"); else { for(int i=1;i<=len;i++) if(dp[i]==Max) ans++; printf("%d %d\n",Max,ans); } return 0; }