# POJ – 2955 Brackets（区间DP）

### Brackets

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 < … < imn, 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

//#include <bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>

using namespace std;
//const int INF = 0x7f7f7f7f;
const int maxn=110;
int dp[maxn][maxn];  //dp[i][j]表示匹配的最大个数

int main()
{
int n;
char str[maxn];
while(~scanf("%s",str+1)){
if(str[1] == 'e')break;
n = strlen(str+1);
for(int i=2;i<=n;i++){ //长度
for(int l=1;l<=n-i+1;l++){  //起点
int r=l+i-1;
dp[l][r]=0;
if((str[l] == '(' && str[r] == ')') || (str[l] == '[' && str[r] == ']')){
dp[l][r]=dp[l+1][r-1] + 2;
}
for(int j=l; j<r; j++){
dp[l][r] = max(dp[l][r], dp[l][j] + dp[j+1][r]);
}
}
}
printf("%d\n", dp[1][n]);
}
return 0;
}