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 … 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

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char s[102];
int dp[102][102];
int main()
{
    int i, j, k, x;
    while(gets(s)!=NULL)
    {
        if(s[0]==‘e‘)break;
        memset(dp,0,sizeof(dp));
        int len= strlen(s);
        for(k=1;k<len;k++) //表示区间长度，从0-len更新
        {
            for(i=0,j=k;j<len;i++,j++)
            {
                if(s[i]==‘(‘&&s[j]==‘)‘||s[i]==‘[‘&&s[j]==‘]‘) //匹配
                    dp[i][j]=dp[i+1][j-1]+2;
                for(x=i;x<j;x++)   //区间最值合并
                    dp[i][j]=max(dp[i][j],dp[i][x]+dp[x+1][j]);
            }
        }
        printf("%d\n",dp[0][len-1]);
    }
    return 0;
}

POJ 2955?Brackets