POJ - 2955 Brackets

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

问符合要求的匹配括号一共有多少。
这里采用逆向思维，将不符合要求的最小值算出来，总数减去结果即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>

using namespace std;

int dp[105][105];
char s[105];

int main()
{
    while(scanf("%s",s),strcmp(s, "end") != 0)
    {
        memset(dp,0,sizeof(dp));
        int len=strlen(s);
        for(int i=0;i<len;i++)
            dp[i][i]=1;
        int j;
        for(int l=1;l<len;l++)
            for(int i=0;i<len-1;i++)
            {
                j=i+l;
                if(j>=len)
                    continue;
                dp[i][j]=1e9;
                if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))
                    dp[i][j]=dp[i+1][j-1];
                for(int k=i;k<j;k++)
                    dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);
                //cout<<i+1<<" "<<j+1<<" "<<dp[i][j]<<endl;
            }
        printf("%d\n",len-dp[0][len-1]);
    }
    
    return 0;
}