POJ 2955 Brackets (interval dp bracket matching)

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3951		Accepted: 2078

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, thenAB 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, ai1ai 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

Title Link: http://poj.org/problem?id=2955

Title: Give a sequence of parentheses, ask the number of valid parentheses in the sequence, if a is legal, then [a], (a) are legal, if A B is legal, AB is also legal

Title Analysis: and POJ 1141 that classic bracket matching similar, the problem is more simple, to find a way to transform the problem, since the maximum number of brackets required to match, we consider adding a minimum of parentheses, so that the entire sequence is legal, so it turned into 1141 that problem, The nature of the maximal matching of the brain-motion analogy binary graph, the maximum match + the maximum independent set = the number of points, obviously to add the least points to make the sequence legal, the least number of points is the | maximum independent set |, we require the original sequence of | maximum Match |, the above is purely yy, the following gives the transfer equation, and 11,411
Dp[i][i] = 1;
Then enumerate the interval lengths
1) Peripheral matching: dp[i][j] = dp[i + 1][j-1];
2) perimeter mismatch, enumeration split point: Dp[i][j] = min (Dp[i][j], dp[i][k] + dp[k + 1][j]); (I <= K < j)

#include <cstdio> #include <cstring> #include <algorithm>using namespace std;int const INF = 0x3fffffff  ; Char s[205];int dp[205][205];int main () {while    (scanf ("%s", s)! = EOF && strcmp (S, "end")! = 0)    {        int Len = strlen (s);        memset (DP, 0, sizeof (DP));        for (int i = 0; i < len; i++)            dp[i][i] = 1;        for (int l = 1, l < len; l++)        {for            (int i = 0; i < len-l; i++)            {                Int j = i + L;                DP[I][J] = INF;                if ((s[i] = = ' (' && s[j] = = ') ') | | (S[i] = = ' [' && s[j] = = '] '))                    DP[I][J] = dp[i + 1][j-1];                for (int k = i; k < J; k++)                    dp[i][j] = min (Dp[i][j], dp[i][k] + dp[k + 1][j]);            }        }        printf ("%d\n", Len-dp[0][len-1]);}    }

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POJ 2955 Brackets (interval dp bracket matching)