Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6585		Accepted: 3534

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

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题意：求最大括号匹配

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f[i][j]表示i到j的最大括号匹配

两种转移

#include
      
       #include
       
        #include
        
         #include
         
          using namespace std;const int N=105;char s[N];int f[N][N];inline bool check(int i,int j){    if(s[i]=='['&&s[j]==']') return 1;    if(s[i]=='('&&s[j]==')') return 1;    return 0;}int main(){    while(~scanf("%s",s+1)){        if(s[1]=='e') break;        memset(f,0,sizeof(f));        int n=strlen(s+1);        for(int i=n;i>=1;i--)            for(int j=i+1;j<=n;j++){                if(check(i,j)) f[i][j]=f[i+1][j-1]+2;                for(int k=i;k<=j;k++) f[i][j]=max(f[i][j],f[i][k]+f[k][j]);             }        printf("%d\n",f[1][n]);            }}