POJ-2955-Brackets【区间DP】

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 of s. That is, you wish to find the largest m such that for indices i1, 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

题目链接：POJ-2955

题目大意：给出括号序列，问匹配的括号有多少

题目思路：区间dp

以下是代码：

#include <iostream>
#include <iomanip>
#include <fstream>
#include <sstream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <functional>
#include <numeric>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <vector>
#include <queue>
#include <deque>
#include <list>

using namespace std;

#define maxn 200
int dp[maxn][maxn];
bool match(char a, char b)
{
if (a == '(' && b == ')') return 1;
if (a == '[' && b == ']') return 1;
return 0;
}

int main()
{
string s;
while(cin >> s)
{
if (s == "end") break;
memset(dp, 0, sizeof dp);

int len = s.size();

for (int i = len - 2; 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 (match(s[i],s[k]))
dp[i][j] = max(dp[i][j], dp[i + 1][k - 1] + dp[k + 1][j] + 2);
}
}
}
cout << dp[0][len - 1] << endl;
}
return 0;
}