Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 6635 | Accepted: 3569 |
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
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
int i,j,k,x,len,dp[102][102];
char a[102];
while(gets(a)!=NULL)
{
if(a[0]=='e') break;
memset(dp,0,sizeof(dp));
len=strlen(a);
for(k=1;k<len;k++)//表示区间长度
for(i=0,j=k;j<len;i++,j++)
{
if(a[i]=='('&&a[j]==')'||a[i]=='['&&a[j]==']')
dp[i][j]=dp[i+1][j-1]+2;//对应两个
for(x=i;x<j;x++)// 区间最值合并
dp[i][j]=max(dp[i][j],dp[i][x]+dp[x+1][j]);
}
printf("%d\n",dp[0][len-1]);
}
return 0;
}