Brackets（区间dp）

Posted 2020-09-27 小时のblog

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

6
6
4
0
6

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[101];
int dp[101][101];
int main()
{
    while(gets(s)!=NULL)
    {
        if(s[0]==‘e‘)break;
        memset(dp,0,sizeof(dp));
        int len=strlen(s);
        for(int i=1;i<=len;i++)
        for(int j=0,k=i;k<=len;j++,k++)
        {
            if(s[j]==‘(‘&&s[k]==‘)‘||s[j]==‘[‘&&s[k]==‘]‘)
            dp[j][k]=dp[j+1][k-1]+2;
            for(int p=j;p<=k;p++)
            dp[j][k]=max(dp[j][k],dp[j][p]+dp[p+1][k]);
        }
        printf("%d\n",dp[0][len-1]);
    }
    return 0;
}