Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8085		Accepted: 4299

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

66406

Source

 

 

 

 

根据规则来进行区间DP

1 #include
      
        2 #include
       
         3 #include
        
          4 #include
         
           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