poj 2955 Brackets

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
题意：求括号匹配的最大数量
区间dp模板题吧。。 今天复习一下 状态转移看代码
ac代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <stack>
#include <string>
#define mt(a) memset(a,0,sizeof(a))
using namespace std;
int dp[105][105];
int main()
{
string s;
while(cin>>s)
{
if(s=="end") break;
mt(dp);
int len=s.size();
for(int l=2;l<=len;l++)
{
for(int i=0;i+l-1<len;i++)
{
int j=i+l-1;
for(int k=i;k<j;k++)
{
if((s[k]=='(' && s[j]==')') || (s[k]=='[' && s[j]==']'))
{
dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j-1]+2);// 区间dp的状态转移是典型的用小区间去推出大区间 我们这里枚举每个长度内可以匹配的最大值
}
else dp[i][j]=max(dp[i][j],max(dp[i][k],dp[k][j]));// 匹配不了就找出子区间的最大值
}
}
}
cout<<dp[0][len-1]<<endl;
}
return 0;
}