Brackets (区间DP)

绝地灬酷狼 2023-06-01 06:11 3阅读 0赞

*  Brackets (区间DP)
    
     *  题目
     *  题意
     *  思路
     *  题解

# Brackets (区间DP) #

## 题目 ##

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 … an, 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, …, im where 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … aim 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 ##

输出能够匹配的最多的括号个数

## 思路 ##

由小区间的数量推出大区间的括号数量

1.  对于小区间 如果能够首尾匹配那么 **dp\[i\]\[j\]=dp\[i+1\]\[j-1\]+2** 因为这是最直接的。
2.  但是这样不是最准确的。
3.  对于 **()()** 来说 **dp\[2\]\[3\]=0** 因为 **)(** 不匹配。那么用一算出来的 **dp\[1\]\[4\]=0+2=2** 是错的，其应该等于 **4** .
4.  所以这就应该进行扫描子区间。 实际最大值应该是  
    **dp\[1\]\[4\] = max( dp\[1\]\[4\] , dp\[1\]\[2\] + dp\[3\]\[4\] ) = 4**

## 题解 ##

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int d[105][105]; int main() { char s[105]; while (scanf("%s", s + 1), s[1] != 'e') { memset(d, 0, sizeof d); int len = strlen(s + 1); //先枚举小区间，因为后面计算大区间时需要使用小区间 for (int l = 0; l <= len; l++) { for (int i = 1; i + l - 1 <= len; i++) { int j = l + i - 1; //基本的状态转移 if ((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']')) { d[i][j] = d[i + 1][j - 1] + 2; } //为了找到真正最大的值，进行扫描。 for (int k = i; k < j; k++) { d[i][j] = max(d[i][j], d[i][k] + d[k + 1][j]); } } } cout << d[1][len] << endl; } }

转载于:https://www.cnblogs.com/tttfu/p/11291003.html