POJ - 2955 Brackets括号匹配(区间dp)
Brackets
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
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=;
char s[N];
int f[N][N];
inline bool check(int i,int j){
if(s[i]=='['&&s[j]==']') return ;
if(s[i]=='('&&s[j]==')') return ;
return ;
}
int main(){
while(~scanf("%s",s+)){
if(s[]=='e') break;
memset(f,,sizeof(f));
int n=strlen(s+);
for(int i=n;i>=;i--)
for(int j=i+;j<=n;j++){
if(check(i,j)) f[i][j]=f[i+][j-]+;
for(int k=i;k<=j;k++) f[i][j]=max(f[i][j],f[i][k]+f[k][j]);
}
printf("%d\n",f[][n]);
}
}
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