POJ 2955 Brackets

Links: http://poj.org/problem?id=2955

Brackets

Time limit:1000ms		Memory limit:65536k
Total submissions:4378		accepted:2331

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] is 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 is regular brackets sequences:

(), [], (()), ()[], ()[()]

While the following character sequences is not:

(, ], )(, ([)], ([(]

Given a brackets sequence of characters a1a2 ... An, your goal are to find the length of the longest regular brackets sequence, which is a subsequence ofs. That's, you wish to find the largest m such this for indicesi1, i2, ..., im where 1≤ i1 < i2 < ... < im ≤ n, ai 1 ai 2 ... aim is a regular brackets sequence.

Given the initial sequence ([([]])] , the longest regular brackets subsequence is [([])] .

Input

The input test file would contain multiple test cases. Each input test case consists of a single line containing only the characters ( , ) , [ , and ] ; St'll has length between 1 and inclusive. The End-of-file is marked by a line containing the word "end" and should not being 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

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

Sample Output

66406

Source

Stanford Local 2004

To give you a sequence of parentheses of not more than 100 length, the length of the longest legal brace subsequence. The valid sequence is determined as: 1. The empty sequence is legal; 2. If a sequence s is legal, then (s) and [s] are both legal and 3. If two sequences A and B are legal, then AB is also legal. For example: (), [], (()), () [], () [()] are legal, and (,],) (, [)], ([[] neither is legal.

Idea--Typical interval DP model. Analysis of the problem can be found: if you find a pair of matching parentheses [xxx]oooo, the interval is divided into two parts, part xxx, the other part is oooo. Set Dp[i][j] represents the longest legal bracket subsequence length between interval [I,j], then dp[i][j]=dp[i+1][j] when i<j, if there is no bracket matching the i in the interval [i+1,j]; If there is a matching K, then dp[i][j] = max{dp[i+1][j],dp[i+1][k-1]+dp[k+1][j]+
2 (I+1<k<j&&i and k are pairs of matching parentheses)}. Therefore, we enumerate I from the end of the string to the end of the string, the last dp[0][len-1] is the answer, Len represents the string length.

Complexity analysis--time complexity: O (len^3), spatial complexity: O (len^2)

Attach the AC code:

#include <iostream> #include <cstdio> #include <string> #include <cmath> #include <iomanip > #include <ctime> #include <climits> #include <cstdlib> #include <cstring> #include < algorithm> #include <queue> #include <vector> #include <set> #include <map>//#pragma comment (Linker, "/stack:102400000, 102400000") using namespace std;typedef unsigned int li;typedef long long ll;typedef unsigned l Ong long Ull;typedef long double ld;const double pi = ACOs ( -1.0); const DOUBLE E = exp (1.0); const double EPS = 1e-8;const i NT MaxLen = 105;char str[maxlen];int dp[maxlen][maxlen];bool match (char A, char b); Whether the parentheses match the int main () {Ios::sync_with_stdio (false), while (scanf ("%s", str) && strcmp ("End", str)) {memset (DP, 0, sizeof (DP)); initialize int len = strlen (str), for (int i=len-1; i>=0; i--) {//From behind widening interval for (int j=i+1; j<len; J + +) {//DP[I][J] = interval [i , j] the longest legal bracket subsequence length dp[i][j] = dp[i+1][j]; The interval [i+1,j] does not match the I bracket for (int k=i+1; k&Lt;=j; k++)//Find to match, split in Split (Match (Str[i], str[k])//compare size, take large dp[i][j] = max (dp[i][j], dp[i+1][k-1]+dp[k+1][j]+2);}} printf ("%d\n", Dp[0][len-1]);} return 0;} BOOL Match (char A, char b) {if ((a== ' (' &&b== ') ') | | (a== ' [' &&b== ']) return 1;return 0;}

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POJ 2955 Brackets