poj 2955 括号匹配 区间dp

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6033		Accepted: 3220

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

Source

Stanford Local 2004

 

题意：给你一个串 问你有多少可匹配的括号  ‘(’   ')'   '['  ']'

 

题解：dp[i][j] 表示 区间i~j有多少可匹配的括号

状态转移方程：dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]) k(为分界点)

但是如果a[i]与a[j]匹配 还需要增加判断 dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2)

来确保最优。

做了一些区间dp，要注意的几点

1.分界点的枚举

2.边界的处理

3.递推过程中的i，j

4.区间dp就是逆着状态递推（dp不都是这样吗 zz）

 

 /******************************
 code by drizzle
 blog: www.cnblogs.com/hsd-/
 ^ ^    ^ ^
  O      O
 ******************************/
 //#include<bits/stdc++.h>
 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<map>
 #include<algorithm>
 #include<cmath>
 #define ll __int64
 #define PI acos(-1.0)
 #define mod 1000000007
 char a[];
 int dp[][];
 using namespace std;
 int main()
 {
     while(gets(a))
     {
         if(a[]=='e')
             break;
         int len=strlen(a);
         memset(dp,,sizeof(dp));
         /*for(int i=; i<len-; i++)
         {
             if(((a[i]=='(')&&(a[i+]==')'))||((a[i]=='[')&&(a[i+]==']')))//边界处理
                 dp[i][i+]=;
         }*/此处删去仍然可以 ac 细细想一下 其实这个边界 已经在下面的if中处理掉了
         for(int i=len-; i>=; i--)
         {
             for(int j=i+; j<=len-; j++)
             {
                 for(int k=i; k<=j; k++)
                     dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+][j]);
                 if(((a[i]=='(')&&(a[j]==')'))||((a[i]=='[')&&(a[j]==']')))
                     dp[i][j]=max(dp[i][j],dp[i+][j-]+);
             }
         }
         cout<<dp[][len-]<<endl;;
     }
     return ;
 }