POJ 2955 Brackets（区间dp）

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

题目大意：

求出满足括号匹配的最长个数

解题思路：

设d[i][j]表示[i,j]满足括号匹配的最长个数，则有

(1)当s[i]与s[j]匹配：

d[i][j]=d[i-1][j-1]+2;

(2)当s[i]与s[j]不匹配：

d[i][j]=max(d[i][k]+d[k+1][j] (k在i+1与j之间）

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char s[105];
int d[105][105];
int main()
{
int n;
cin>>n;
getchar();
while(n--)
{
gets(s);
memset(d,0,sizeof(d));
int len=strlen(s);
for(int k=1;k<len;k++)
{
for(int i=0,j=k;j<len;i++,j++)
{
if(s[i]=='('&&s[j]==')'||s[i]=='['&&s[j]==']')
d[i][j]=d[i+1][j-1]+2;
for(int x=i;x<j;x++)
d[i][j]=max(d[i][j],d[i][x]+d[x+1][j]);
}
}
cout<<len-d[0][len-1]<<endl;
}
return 0;
}