POJ2955——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 ofs. That is, you wish to find the largest m such that for indicesi₁, 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[([])].

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.

For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.

#include <map>  
#include <set>  
#include <list>  
#include <stack>  
#include <queue>  
#include <vector>  
#include <cmath>  
#include <cstdio>  
#include <cstring>  
#include <iostream>  
#include <algorithm>  
  
using namespace std;

char str[110];
int dp[110][110];

int main()
{
	while (~scanf("%s", str), str[0] != 'e')
	{
		int len = strlen(str);
		memset (dp, 0, sizeof(dp));
		for (int i = len - 1; i >= 0; --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 ((str[i] == '(' && str[k] == ')') || (str[i] == '[' && str[k] == ']'))
					{
						dp[i][j] = max(dp[i][j], dp[i + 1][k - 1] + dp[k + 1][j] + 2);
					}
				}
			}
		}
		printf("%d\n", dp[0][len - 1]);
	}
	return 0;
}