POJ-2955 Brackets （区间DP）

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9363 Accepted: 5006

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

66406

#include <stdio.h>#include <string.h>#include <iostream>using namespace std;int dp[101][101];string s;int main(){while(cin >> s){if(s[0] == 'e') return 0;memset(dp, 0, sizeof(dp));for(int d = 1; d <= s.size(); ++d){for(int i = 0; i + d < s.size(); ++i){if(s[i] == '(' && s[i + d] == ')'){dp[i][i + d] = dp[i + 1][i + d - 1] + 2;}if(s[i] == '[' && s[i + d] == ']'){dp[i][i + d] = dp[i + 1][i + d - 1] + 2;}for(int j = i; j < i + d; ++j){dp[i][i + d] = max(dp[i][i + d], dp[i][j] + dp[j + 1][i + d]);}}}printf("%d\n", dp[0][s.size() - 1]);}}/*题意：问一个括号序列中最长的合法括号子序列。思路：区间dp，dp[i][j]表示[i,j]子串中最长的合法括号子序列是多长。按照括号合法的规则去转移一下就好了。*/