POJ 2955- Brackets[区间dp]

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

思路: 对左括号不进行处理, rev数组存储一个括号的反括号 如rev['('] = ')' 其余的同理

         dp[i][j] 代表的是i到j位置匹配的最大个数

         转移方程dp[i][j] = dp[i][j-1], 对任意i,j

                       dp[i][j] = (dp[i][k] + dp[k+1][j] +1, a[k] = rev[a[j]]

        answer = dp[0][N-1] *2

#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<iomanip>
#include<stdlib.h>
#include<cstdio>
#include<string>
#include<string.h>
#include<set>
#include<stack>
#include<map>
using namespace std;

#define rep(i, n) for(int i=0; i<n ;i++)
#define rep1(i, n) for(int i=1; i<=n; i++)
typedef long long ll;
typedef  pair<int,int> P;
const int INF = 0x7fffffff;
const int MAX_N = 105;
const int MAX_V = 0;
const int MAX_M = 0;
const int MAX_Q = 1e5+5;

int  dp[MAX_N][MAX_N];
int l[3] = {'(','[','{'}, r[3] = {')',']','}'};
int rev[256];

int main(){
string a;
for(int i=0; i<3; i++){
rev[l[i]] = r[i];
rev[r[i]] = l[i];
}
while(getline(cin, a)){
if(a=="end") break;
int N = a.length();
for(int i=0; i<N; i++){
dp[i][i] = 0;
}
for(int j=1; j<N; j++){
for(int i=0; i<j; i++){
dp[i][j] = dp[i][j-1];
bool flag = false;
for(int s =0; s<3; s++)
if(a[j]==r[s])
flag = true;
if(flag){
for(int k=i; k<j; k++){
if(rev[a[k]]==a[j]){
dp[i][j] = max(dp[i][j], dp[i][k]+dp[k+1][j-1]+1);
}
}
}
}
}
cout<<dp[0][N-1]*2<<endl;
}
}