POJ 2955 Brackets

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7607		Accepted: 4043

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

找到拥有括号层数最多的以堆括号有多少个括号，()2个，([)]2个，([])]4个，([])]]]]4个，((()))6个。

区间就是第i个和第i+k之间的遍历一遍，k就代表区间大小。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s[102];
int dp[102][102];
int main()
{
int i,j,k,n;
while(~scanf("%s",s))
{
if(s[0]=='e')
break;
n=strlen(s);
memset(dp,0,sizeof(dp));
for(k=1;k<n;k++)
{
for(i=0;i+k<n;i++)
{
if(s[i]=='('&&s[i+k]==')'||s[i]=='['&&s[i+k]==']')
dp[i][i+k]=dp[i+1][i+k-1]+2;
for(j=i;j<i+k;j++)
{
dp[i][i+k]=max(dp[i][i+k],dp[i][j]+dp[j+1][i+k]);
}
}
}
printf("%d\n",dp[0][n-1]);
memset(s,0,sizeof(s));
}
return 0;
}