Brackets(区间dp）

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3624		Accepted: 1879

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 a₁a₂ … a_n, 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, …, i_mwhere 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im 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

6
6
4
0
6

Source

Stanford Local 2004

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
char st[150] ;
int a[110][110] ;

bool check (int a , int b)
{
    if (st[a] == '(' && st[b] == ')' )
        return 1 ;
    if (st[a] == '[' && st[b] == ']' )
        return 1 ;
    return 0 ;
}

int main ()
{
   // freopen ("a.txt" , "r" , stdin ) ;
    while (1) {
        gets (st) ;
        if (strcmp (st , "end") == 0)
            break ;
        memset (a , 0 , sizeof(a) ) ;
        int len = strlen (st) ;
        for (int o = 2 ; o <= len ; o++) {
            for (int i = 0 ;  i < len - o + 1; i++) {
                int j = i + o ;
                for (int k = i ; k < j ; k++ ) {
                    a[i][j - 1] = max (a[i][j - 1] , a[i][k] + a[k + 1][j - 1] ) ;
                    if (check (i , j - 1) ) {
                        a[i][j - 1] = max (a[i][j - 1] , a[i + 1][j - 1 - 1] + 2 ) ;
                    }
                }
            }
        }
        printf ("%d\n" , a[0][len - 1] ) ;
    }
    return 0 ;
}

区间dp感觉和merge sort有异曲同工之妙
从可行的最小区间出发，逐级上去，最终得到整段区间的最终结。

dp[i][j] 指[i , j]这段区间的最优解