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Brackets
阅读量:6910 次
发布时间:2019-06-27

本文共 1861 字,大约阅读时间需要 6 分钟。

Brackets
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 3871   Accepted: 2028

Description

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 a1a2an, 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 < … < imn, ai1ai2aim 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

66406
1 #include
2 #include
3 #define max(x,y) x>y?x:y 4 int main(){
char sequence[110]; 5 int dp[105][105],t; 6 while(scanf("%s",sequence),strcmp(sequence,"end")){ 7 memset(dp,0,sizeof(dp)); 8 t=strlen(sequence); 9 for(int i=t-2;i>=0;i--){10 for(int j=i+1;j

 

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