We give the following inductive definition of a “regular brackets” sequence:
the empty sequence is a regular brackets sequence, if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and if a and b are regular brackets sequences, then ab is a regular brackets sequence. no other sequence is a regular brackets sequence For instance, all of the following character sequences are regular brackets sequences:
(), [], (()), ()[], ()[()]
while the following character sequences are not:
(, ], )(, ([)], ([(]
Given a brackets sequence of characters a1a2 … an, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such that for indices i1, i2, …, im where 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … aim is a regular brackets sequence.
Given the initial sequence ([([]])], the longest regular brackets subsequence is [([])].
Input The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters (, ), [, and ]; each input test will have length between 1 and 100, inclusive. The end-of-file is marked by a line containing the word “end” and should not be processed.
Output For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.
Sample Input ((())) ()()() ([]]) )[)( ([][][) end Sample Output 6 6 4 0 6
#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> using namespace std; char s[100000]; int dp[110][110]; int main() { while(gets(s+1)) { if(s[1]=='e') break; int n=strlen(s+1); memset(dp,0,sizeof(dp)); for(int len=2;len<=n;len++) for(int i=1;i+len-1<=n;i++) { int l=i,r=i+len-1; if(s[l]=='('&&s[r]==')'||s[l]=='['&&s[r]==']') dp[l][r]=dp[l+1][r-1] +2; for(int k=l;k<r;k++) dp[l][r]=max(dp[l][r],dp[l][k]+dp[k+1][r]); } cout<<dp[1][n]<<endl; } }