Brackets（区间dp）

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8017		Accepted: 4257

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

Source

Stanford Local 2004

【题目大意】

最大括号匹配

【思路】

区间dp 枚举长度

【code】

#include<iostream>

#include<cstdio>

#include<cstring>

using namespace std;

char s[];

int dp[][];

int main()

{

    while(gets(s)!=NULL)

    {

        if(s[]=='e')break;

        memset(dp,,sizeof(dp));

        int len=strlen(s);

        for(int i=;i<=len;i++)

        for(int j=,k=i;k<=len;j++,k++)

        {

            if(s[j]=='('&&s[k]==')'||s[j]=='['&&s[k]==']')

            dp[j][k]=dp[j+][k-]+;

            for(int p=j;p<=k;p++)

            dp[j][k]=max(dp[j][k],dp[j][p]+dp[p+][k]);

        }

        printf("%d\n",dp[][len-]);

    }

    return ;

}

巴特西

Brackets（区间dp）

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