Brackets
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 8017
Accepted: 4257
Description
We give the following inductive definition of 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 a_1_a_2 … _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 i_1, _i_2, …, _im where 1 ≤i_1 < _i_2 < … < _im ≤ n, ai_1_ai_2 … _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
【题目大意】
最大括号匹配
【思路】
区间dp 枚举长度
【code】
#include
#include
#include
using namespace std;
char s[];
int dp[][];
int main()
{
while(gets(s)!=NULL)
{
if(s[]=='e')break;
memset(dp,,sizeof(dp));
int len=strlen(s);
for(int i=;i<=len;i++)
for(int j=,k=i;k<=len;j++,k++)
{
if(s[j]=='('&&s[k]==')'||s[j]=='['&&s[k]==']')
dp[j][k]=dp[j+][k-]+;
for(int p=j;p<=k;p++)
dp[j][k]=max(dp[j][k],dp[j][p]+dp[p+][k]);
}
printf("%d\n",dp[][len-]);
}
return ;
}
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