poj-2955 (区间dp)

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
6AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
char str[150];
int dp[120][120];
int main()
{
    while(1)
    {
        scanf("%s",str);
        if(str[0]==‘e‘)break;
    int len=strlen(str);
    memset(dp,0,sizeof(dp));
    for(int j=0;j<len;j++)
    {
        for(int i=j-1;i>=0;i--)
        {
            dp[i][j]=dp[i+1][j];
            for(int k=i+1;k<=j;k++)
            {
                if((str[i]==‘(‘&&str[k]==‘)‘)||(str[i]==‘[‘&&str[k]==‘]‘))
                   {
                       dp[i][j]=max(dp[i][j],dp[i+1][k-1]+dp[k+1][j]+2);
                   }
            }
        }
    }
    cout<<dp[0][len-1]<<"\n";
    }
    return 0;
}