題解 print ini between have indices scrip -s break

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_m where 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
題解
區間dp，和1141那道題做法很像

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 using namespace std;
 5 int dp[110][110];
 6 char str[110];
 7 bool ok(int x,int y)
 8 {
 9     if(str[x]==‘(‘&&str[y]==‘)‘)return true;
10     if(str[x]==‘[‘&&str[y]==‘]‘)return true;
11     return false;
12 }
13 int main()
14 {
15     while(1)
16     {
17         scanf("%s",str+1);
18         if(str[1]==‘e‘)break;
19         int l=strlen(str+1);
20         for(int k=2 ; k<=l ; ++k )
21             for(int i=1,j=i+k-1 ; j<=l ; ++i,j=i+k-1)
22                 {
23                     dp[i][j]=0;
24                     if(ok(i,j))dp[i][j]=dp[i+1][j-1]+2;
25                     for(int p=i ; p<j ; ++p )
26                         dp[i][j]=max(dp[i][j],dp[i][p]+dp[p+1][j]);    
27                 }
28         printf("%d\n",dp[1][l]);
29     }
30     return 0;
31 }

poj2955：Brackets