poj 2955 Brackets

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1734		Accepted: 869

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

Source

Stanford Local 2004

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，一开始自己还以为自己套lrj书上的思路，后来才发现两者间的不同，之间差别就是此题需要把子结构都遍历一次，本来想看看有什么更快的，开始找到的都更这个差不多思路

状态转移方程：if(mach(i,j))f[i][j]=f[i+1][j-1]+1;

f[i][j]=max(f[i][j],f[i][i+g]+f[i+g+1][j])《0<=g<k》

#include <stdio.h>
#include <string.h>
int f[110][110];
char s[110];
int max(int a,int b){return a>b?a:b;}
int mach(int a,int b){if((s[a]=='['&&s[b]==']')||(s[a]=='('&&s[b]==')'))return 1;return 0;}
int main()
{
int i,j,k,len,g;
while(scanf("%s",s)&&s[0]!='e')
{
memset(f,0,sizeof(f));
len=strlen(s);
for(i=0;i<len;i++){
f[i][i]=0;
if(mach(i,i+1))f[i][i+1]=1;
}
for(k=2;k<len;k++)
for(i=0;i<len-k;i++)
{
j=i+k;
if(mach(i,j))f[i][j]=f[i+1][j-1]+1;
for(g=0;g<k;g++)
f[i][j]=max(f[i][j],f[i][i+g]+f[i+g+1][j]);
}
printf("%d\n",f[0][len-1]*2);
}
return 0;
}