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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5150 | Accepted: 2761 |
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 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
AC代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
char str[150];
int dp[120][120];
int main()
{
while(1)
{
scanf("%s",str);
if(str[0]==‘e‘)break;
int len=strlen(str);
memset(dp,0,sizeof(dp));
for(int j=0;j<len;j++)
{
for(int i=j-1;i>=0;i--)
{
dp[i][j]=dp[i+1][j];
for(int k=i+1;k<=j;k++)
{
if((str[i]==‘(‘&&str[k]==‘)‘)||(str[i]==‘[‘&&str[k]==‘]‘))
{
dp[i][j]=max(dp[i][j],dp[i+1][k-1]+dp[k+1][j]+2);
}
}
}
}
cout<<dp[0][len-1]<<"\n";
}
return 0;
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原文地址：http://www.cnblogs.com/zhangchengc919/p/5246880.html
