POJ 2955 Brackets, poj2955brackets

POJ 2955 Brackets, poj2955brackets
Brackets

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:6622		Accepted:3558

Description

We give the following inductive definition of a "regular brackets" sequence:

• The empty sequence is a regular brackets sequence,
• IfSIs a regular brackets sequence, then (S) And [S] Are regular brackets sequences, and
• IfAAndBAre regular brackets sequences, thenABIs 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 charactersA1A2...An, Your goal is to find the length of the longest regular brackets sequence that is a subsequenceS. That is, you wish to find the largestMSuch that for indicesI1,I2 ,...,ImWhere 1 ≤I1 <I2 <... <Im≤N,Ai1Ai2...AimIs 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, intrusive. The end-of-file is marked by a line containing the word "end" and shocould not be processed.

Output

For each input case, the program shocould print the length of the longest possible regular brackets subsequence on a single line.

Sample Input

((()))()()()([]]))[)(([][][)end

Sample Output

66406

SourceStanford Local 2004 question: give you a sequence of parentheses whose length cannot exceed 100, and find the length of the longest legal sequence of parentheses. A valid sequence of parentheses meets the following conditions: 1. the null sequence of parentheses is valid. 2. if a bracket sequence s is valid, both (s) and [s] are valid. 3. if the two parentheses a and B are valid, AB is also valid. 4. other parentheses are invalid. For example: (), [], (), () [], () [()] are legal, and (,],) (, ([)], ([(] is invalid. Solution: a typical topic of the interval DP model. After analyzing the problem, we can find that if we find a pair of matching parentheses, such as [xxx] oooo, we divide the interval into two parts: xxx and oooo. Set dp [I] [j] to indicate the length of the longest legal parentheses subsequence between intervals [I, j]. When I <j, if the interval [I + 1, if no parentheses match with I exist in j, dp [I] [j] = dp [I + 1] [j, so dp [I] [j] = max {dp [I + 1] [j], dp [I + 1] [k-1] + dp [k + 1] [j] + 1 (I <= k <= j & I and k are a pair of matching brackets) }. Therefore, we use the length of the entire string as the interval for search, then the last 2 * dp [0] [len-1] is the answer, len represents the length of the string. For details, see the code. Attach the AC code: 1 # include <cstdio> 2 # include <cstring> 3 # include <algorithm> 4 using namespace std; 5 const int maxn = 105; 6 char str [maxn]; 7 int dp [maxn] [maxn]; 8 9 bool match (char a, char B) {10 return (a = '(' & B = ') | (a =' ['& B ='] '); 11} 12 13 int dfs (int l, int r) {14 if (l> r) 15 return 0; 16 if (l = r) 17 return dp [l] [r] = 0; 18 if (l + 1 = r) 19 return dp [l] [r] = match (str [l], str [r]); 20 if (dp [L] [r]! =-1) 21 return dp [l] [r]; 22 int ans = dfs (l + 1, r); 23 for (int I = l; I <= r; + + I) 24 if (match (str [l], str [I]) 25 ans = max (ans, dfs (l + 1, I-1) + dfs (I + 1, r) + 1); 26 return dp [l] [r] = ans; 27} 28 29 int main () {30 while (~ Scanf ("% s", str) & str [0]! = 'E') {31 memset (dp,-1, sizeof (dp); 32 int len = strlen (str); 33 dfs (0, len-1 ); 34 printf ("% d \ n", 2 * dp [0] [len-1]); 35} 36 return 0; 37}View Code