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 6 4 0 6
题意:类似于最长公共子序列的最大匹配括号数
思路:明显就是一种自底向上型DP,
// 求最长合法子序列
// 区间 dp
// dp[i][j] 表示区间 i,j 的最长合法子序列
#include <iostream> #include <string> #include <algorithm> using namespace std; int dp[105][105]; bool cmp(char c1, char c2) { if(c1 == '(' && c2 == ')') return true; if(c1 == '[' && c2 == ']') return true; return false; } int main() { strings; while(cin>>s && s != "end") { int length = s.length(); for(int i = 0 ; i < length; i++) { dp[i][i] = 0; if(cmp(s[i],s[i+1])) { dp[i][i+1] = 2; } else dp[i][i+1] = 0; } for(int k=3;k<=length;k++) { for(int i=0;i+k-1<length;i++) { dp[i][i+k-1]=0; if(cmp(s[i],s[i+k-1])) { dp[i][i+k-1]=dp[i+1][i+k-2]+2; } for(int j=i;j<i+k-1;j++) { dp[i][i+k-1]=max(dp[i][i+k-1],dp[i][j]+dp[j+1][i+k-1]); } } } cout<<dp[0][length-1]<<endl; } return 0; }