Poj 2955 brackets (motion gauge)

Brackets

Time limit:1000 ms		Memory limit:65536 K
Total submissions:2999		Accepted:1536

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

Source

Stanford local 2004
? If a pair of matching brackets [XXX] Oooo is found, the interval is divided into two parts: XXX and oooo. Then, recursion is performed until the Interval Length is 1 or 2? DP [I] [J] indicates the maximum length of the Legal parentheses subsequence between [I, j? DP [I] [J] = min {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)}
Code:

# Include <iostream> # include <string. h> # include <math. h> # include <stdio. h> using namespace STD; # define M 110int DP [m] [m]; char STR [m]; int math (int x, int y) // determine whether to pair. {Return STR [x] = '(' & STR [y] = ') '| STR [x] =' {'& STR [y] ='} '| STR [x] =' ['& STR [y] = ']';} int DFS (int s, int e) {int I, RET, T; If (E <s) return 0; // consider various special cases. If (E = s) return DP [s] [e] = 0; If (E-S = 1) return DP [s] [e] = math (S, e); If (DP [s] [e]! =-1) return DP [s] [E]; // if this status has been calculated. Ret = DFS (S + 1, e); // The posture is incorrect. You can still look at other people's. For (I = S + 1; I <= E; I ++) // find matching parentheses. If (Math (S, I) {T = DFS (S + 1, I-1) + DFS (I + 1, E) + 1; if (T> RET) ret = T;} return DP [s] [e] = ret;} int main () {int I, j, k, n; while (scanf ("% s ", str) {If (STR [0] = 'E') break; n = strlen (STR); memset (DP,-1, sizeof (DP )); DFS (0, n-1); printf ("% d \ n", DP [0] [n-1] * 2);} return 0 ;}