Brackets

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 … an, 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, …, im where 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … aim 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

根据规则来进行区间DP

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #define lli long long int 
 6 using namespace std;
 7 const int MAXN=1001;
 8 const int maxn=0x7fffff;
 9 void read(int &n)
10 {
11     char c='+';int x=0;bool flag=0;
12     while(c<'0'||c>'9'){c=getchar();if(c=='-')flag=1;}
13     while(c>='0'&&c<='9')
14     x=(x<<1)+(x<<3)+c-48,c=getchar();
15     flag==1?n=-x:n=x;
16 }
17 char s[MAXN];
18 int dp[MAXN][MAXN];
19 int  main()
20 {
21     while(scanf("%s",s))
22     {
23         if(s[0]=='e')
24             break;
25         memset(dp,0,sizeof(dp));
26         int l=strlen(s);
27         for(int i=l;i>=0;i--)
28             for(int j=i;j<l;j++)
29             {
30                 //dp[i][j]=max(dp[i+1][j],dp[i][j-1]);
31                 
32                 if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))
33                     dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2);
34                     
35                 for(int k=i;k<j;k++)    
36                     dp[i][j]=max(dp[i][k]+dp[k+1][j],dp[i][j]);
37             
38             }
39         printf("%d\n",dp[0][l-1]);
40     }
41     return 0;
42 }