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[105];
int dp[105][105];
using namespace std;
int main()
{
    while(gets(a))
    {
        if(a[0]=='e')
            break;
        int len=strlen(a);
        memset(dp,0,sizeof(dp));
        /*for(int i=0; i<len-1; i++)
        {
            if(((a[i]=='(')&&(a[i+1]==')'))||((a[i]=='[')&&(a[i+1]==']')))//边界处理
                dp[i][i+1]=2;
        }*/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-1; i>=0; i--)
        {
            for(int j=i+1; j<=len-1; j++)
            {
                for(int k=i; k<=j; k++)
                    dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
                if(((a[i]=='(')&&(a[j]==')'))||((a[i]=='[')&&(a[j]==']')))
                    dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2);
            }
        }
        cout<<dp[0][len-1]<<endl;;
    }
    return 0;
}