poj2955：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
题解
区间dp，和1141那道题做法很像

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int dp[110][110];
char str[110];
bool ok(int x,int y)
{
    if(str[x]=='('&&str[y]==')')return true;
    if(str[x]=='['&&str[y]==']')return true;
    return false;
}
int main()
{
    while(1)
    {
        scanf("%s",str+1);
        if(str[1]=='e')break;
        int l=strlen(str+1);
        for(int k=2 ; k<=l ; ++k )
            for(int i=1,j=i+k-1 ; j<=l ; ++i,j=i+k-1)
                {
                    dp[i][j]=0;
                    if(ok(i,j))dp[i][j]=dp[i+1][j-1]+2;
                    for(int p=i ; p<j ; ++p )
                        dp[i][j]=max(dp[i][j],dp[i][p]+dp[p+1][j]);    
                }
        printf("%d
",dp[1][l]);
    }
    return 0;
}