| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: | Accepted: |
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
代码中详解！！！
AC代码如下：
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[105][105];
char str[1005];
int main()
{
int i,j,o,t;
while(cin>>str,str[0]!='e')
{
int l=strlen (str);
for(i=0;i<l;i++)
for(j=0;j<l;j++)
dp[i][j]=0;//初始化状态
for(o=2;o<=l;o++)//枚举所有区间长度
{
for(i=0;i<l-o+1;i++)
{
j=i+o-1;//控制[i,j]区间长度为o。
for(t=i;t<j;t++)//检索[i,j]中的满足条件的子区间
{
dp[i][j]=max(dp[i][j],dp[i][t]+dp[t+1][j]);//更新子区间合成的区间最大值MAXXX
if((str[i]=='('&&str[j]==')')||(str[i]=='['&&str[j]==']'))
dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2);//比较MAXXX与整体区间的相比谁大
}
}
}
cout<<dp[0][l-1]<<endl;
}
return 0;
}
POJ 2955 Brackets,布布扣,bubuko.com
原文地址：http://blog.csdn.net/hanhai768/article/details/37728699
