poj2955 Brackets--最大括号匹配数

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

最基础的区间dp。dp[i][j]表示 i~j 区间内的最大匹配数。首先假设s[i]和s[j]不匹配,s[i][j]=s[i+1][j],然后在区间k(i+1~j)之间寻找s[i]和s[k]匹配，状态转移 dp[i][j]=max(dp[i][j],dp[i][k-1]+dp[k+1][j]+2);

#include
#include
#include
#include
#include
#include
using namespace std;
int dp[110][110];
char s[110];
int main()
{
    while(scanf("%s",s+1)==1 && s[1]!='e')
    {
        int len = strlen(s+1);
        memset(dp,0,sizeof(dp));
        for(int i=len-1;i>=1;i--)
            for(int j=i+1;j<=len;j++)
            {
                dp[i][j]=dp[i+1][j];
                for(int k=i+1;k<=j;k++)
                {
                    if(s[i]=='('&&s[k]==')' || s[i]=='['&&s[k]==']')
                        dp[i][j] = max( dp[i][j], dp[i+1][k-1]+dp[k+1][j]+2);
                }
            }
        printf("%d\n",dp[1][len]);
    }
}