目录
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
输出能够匹配的最多的括号个数
由小区间的数量推出大区间的括号数量
#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;
}
}
原文:https://www.cnblogs.com/tttfu/p/11291003.html