poj 2955 Brackets

Posted 2020-10-04 不搞事情和咸鱼有什么区别

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：求括号匹配的最大数量
区间dp模板题吧。。 今天复习一下 状态转移看代码
ac代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <stack>
#include <string>
#define mt(a) memset(a,0,sizeof(a))
using namespace std;
int dp[105][105];
int main()
{
    string s;
    while(cin>>s)
    {
        if(s=="end") break;
        mt(dp);
        int len=s.size();
        for(int l=2;l<=len;l++)
        {
            for(int i=0;i+l-1<len;i++)
            {
                int j=i+l-1;
                for(int k=i;k<j;k++)
                {
                    if((s[k]==‘(‘ && s[j]==‘)‘) || (s[k]==‘[‘ && s[j]==‘]‘))
                    {
                        dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j-1]+2);// 区间dp的状态转移是典型的用小区间去推出大区间 我们这里枚举每个长度内可以匹配的最大值
                    }
                    else dp[i][j]=max(dp[i][j],max(dp[i][k],dp[k][j]));// 匹配不了就找出子区间的最大值
                }
            }
        }
        cout<<dp[0][len-1]<<endl;
    }
    return 0;
}