poj 2955 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
题意：类似于最长公共子序列的最大匹配括号数
 思路：明显就是一种自底向上型DP，

状态转移方程：if(mach(i,j))f[i][j]=f[i+1][j-1]+1; f[i][j]=max(f[i][j],f[i][i+g]+f[i+g+1][j])《0<=g<k》
// 求最长合法子序列
// 区间 dp
// dp[i][j] 表示区间 i,j 的最长合法子序列

#include <iostream>
#include <string>
#include <algorithm>
using  namespace std;
int dp[105][105];
bool cmp(char c1, char c2)
{
    if(c1 == '(' && c2 == ')')
     return true;
     if(c1 == '[' && c2 == ']')
     return true;
     return false;
}
int main()
{
    strings;
    while(cin>>s && s != "end")
    {
         int length = s.length();
         for(int i = 0 ; i < length; i++)
         {
             dp[i][i] = 0;
             if(cmp(s[i],s[i+1]))
             {
                 dp[i][i+1] = 2;
             }
             else
             dp[i][i+1] = 0;
         }
        for(int k=3;k<=length;k++)
        {
            for(int i=0;i+k-1<length;i++)
            {
                 dp[i][i+k-1]=0;
                 if(cmp(s[i],s[i+k-1]))
                 {
                     dp[i][i+k-1]=dp[i+1][i+k-2]+2;
                 }

                 for(int j=i;j<i+k-1;j++)
                 {
                      dp[i][i+k-1]=max(dp[i][i+k-1],dp[i][j]+dp[j+1][i+k-1]);
                 }

            }

         }
         cout<<dp[0][length-1]<<endl;
    }
    return 0;

}