POJ 2955 Brackets （区间dp入门）

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

6
6
4
0
6

题意给你一个只含()[]的字符串，问你最多能配成对的有多少个和字符。
区间dp的入门题。整理下思路dp[i][j]表示区间i~j之间最大的匹配字符数。
if ((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']')) ————>dp[i][j]=dp[i+1][j-1]+2; 懂吧
代码如下：

 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 #include <iostream>

 using namespace std;
 char s[];
 int dp[][];
 int main()
 {
     //freopen("de.txt","r",stdin);
     while (~scanf("%s",&s))
     {
         if (s[]=='e')
         break ;
         memset(dp,,sizeof dp);
         int len=strlen(s);
         for (int k=;k<len;++k)
         {
             for (int i=,j=k;j<len;++i,++j)
             {
                 if ((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))
                     dp[i][j]=dp[i+][j-]+;
                 for (int x=i;x<j;x++)
                     dp[i][j]=max(dp[i][j],dp[i][x]+dp[x+][j]);
             }
         }
         printf("%d\n",dp[][len-]);
     }
     return ;
 }