Brackets
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 17043 | Accepted: 8831 |
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 … 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
4
0
6
Source
Solution
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 110;
int dp[maxn][maxn];
string s;
bool match(int i, int j)
{
if (s[i] == '[' && s[j] == ']') return true;
if (s[i] == '(' && s[j] == ')') return true;
return false;
}
int main()
{
while (cin >> s && s != "end")
{
memset(dp, 0, sizeof(dp));
int len = (int)s.length();
for (int l = 1; l < len; l++)
{
for (int i = 0, j = i + l; j < len; i++, j++)
{
if (match(i, j)) dp[i][j] = dp[i + 1][j - 1] + 2;
for (int k = i; k < j; k++) dp[i][j] = max(dp[i][j], dp[i][k] + dp[k + 1][j]);
}
}
printf("%d\n", dp[0][len - 1]);
}
return 0;
}