Brackets (区间DP)

Brackets

 POJ - 2955 

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
题意：
　　给出一个字符串，问该字符串中括号的最大的匹配是多少（）为一对括号的匹配，[]为一对括号的匹配。
题解：
　　区间DP
　　dp[i][j]表示i-j这个区间中括号的最大匹配数，可知对于每一个区间来说，初值为如果s[i] s[j]恰好可以匹配，那么dp[i][j]=dp[i+1][j-1]+2,如果不匹配，那么dp[i][j]=dp[i+1][j-1];
然后遍历区间断点K，求最大值。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s[110];
int dp[110][110];
int main()
{
    while(~scanf("%s",s))
    {
        if(s[0]=='e')
            break;
        memset(dp,0,sizeof(dp));
        int n=strlen(s);
        for(int len=1;len<=n;len++)
        {
            for(int l=0;l+len<n;l++)
            {
                int r=l+len;
                if((s[l]=='('&&s[r]==')')||(s[l]=='['&&s[r]==']'))
                    dp[l][r]=dp[l+1][r-1]+2;
                for(int k=l;k<r;k++)
                    dp[l][r]=max(dp[l][r],dp[l][k]+dp[k+1][r]);
            }
        }
        printf("%d
",dp[0][n-1]);
    }
}