POJ 2955 Brackets （区间dp入门）

Posted 2020-08-17 抓不住Jerry的Tom

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：

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <iostream>
 6 
 7 using namespace std;
 8 char s[110];
 9 int dp[110][110];
10 int main()
11 {
12     //freopen("de.txt","r",stdin);
13     while (~scanf("%s",&s))
14     {
15         if (s[0]==‘e‘)
16         break ;
17         memset(dp,0,sizeof dp);
18         int len=strlen(s);
19         for (int k=1;k<len;++k)
20         {
21             for (int i=0,j=k;j<len;++i,++j)
22             {
23                 if ((s[i]==‘(‘&&s[j]==‘)‘)||(s[i]==‘[‘&&s[j]==‘]‘))
24                     dp[i][j]=dp[i+1][j-1]+2;
25                 for (int x=i;x<j;x++)
26                     dp[i][j]=max(dp[i][j],dp[i][x]+dp[x+1][j]);
27             }
28         }
29         printf("%d\n",dp[0][len-1]);
30     }
31     return 0;
32 }