POJ2955——Brackets

Brackets
Time Limit: 1000MS
 
Memory Limit: 65536K
Total Submissions: 3341
 
Accepted: 1717
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 …an, 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, …, im where 1 ≤i1 < i2 < … < im ≤ n, ai1ai2 …aim 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

Source
Stanford Local 2004
또한 매우 고전적인 구간 dp문제입니다. 우리는 dp[i][j]로 i에서 j까지 최대 괄호 일치수를 나타냅니다.
만약 i번째 괄호가 [i+1, j]에서 일치하지 않는다면 dp[i][j]=dp[i+1][j];
그렇지 않으면 구간 [i, j]에서 k를 찾아 i와 k를 맞추면 구간은 2단, [i+1, k-1]와 [k+1, j]로 나뉜다.
그래서 dp[i][j]=max(dp[i+1][j], dp[i+1][k-1]+dp[k+1][j]+2)

#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
", dp[0][len - 1]);
	}
	return 0;
}