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 200int 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;}