POJ 2955：Brackets（区间DP）

Brackets

Time limit:20000 ms Memory limit:65536 kB

Problem 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：

给出一个括号串，求最长的合法的子序列的长度。

解题思路：

设dp[i][j]为最大长度，那么如果第i个和第j个配对，那么dp[i][j]=dp[i+1][j−1]+2再把所有的dp[i][i+k]+dp[i+k+1][j]都遍历一次，看有没有更大的情况，如果有就更新咯。

Code：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <string>
#include <cstdio>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;

const int maxn=105;
int dp[maxn][maxn];

int main()
{
string str;
while(cin>>str)
{
if(str=="end")
break;
int len=str.length();
mem(dp,0);
for(int i=1;i<len;i++)
{
for(int j=0;j<len-i;j++)
{
if((str[j]=='('&&str[j+i]==')')||str[j]=='['&&str[j+i]==']')
dp[j][j+i]=dp[j+1][j+i-1]+2;
for(int k=0;k<i;k++)
{
dp[j][j+i]=max(dp[j][j+i],dp[j][j+k]+dp[j+k+1][j+i]);
}
}
}
cout<<dp[0][len-1]<<endl;

}
return 0;
}