POJ - 2955 Brackets （区间DP）

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：给你一堆括号，要你求出最大匹配数，经典的括号匹配DP问题。

解题思路：区间DP问题，首先枚举区间，根据题目第二个条件，如果符合则匹配数+2，根据题目第三个条件还要进行中间分隔，枚举所有情况。dp[i][j]代表i到j这个区间的最大匹配数，那么如果s[i]==s[j]，dp[i][j]=dp[i+1][j-1]+2，然后我们再枚举中间截断的情况，

dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j])，详见代码。

#include<iostream>
#include<memory.h>
#include<string>
using namespace std;

string str;
int dp[105][105];//i到j这个区间的最大匹配数

int main(){

while(cin>>str){
if(str=="end")
break;

memset(dp,0,sizeof(dp));

//枚举区间长度，必须要从小到大枚举
for(int len=2;len<=str.size();len++){

//l,r为左右标记，两个标记不断地右移
for(int l=0,r=len-1;r<str.size();l++,r++){

//题目第二个条件
if((str[l]=='('&&str[r]==')')||(str[l]=='['&&str[r]==']'))
dp[l][r]=dp[l+1][r-1]+2;

//题目第三个条件，枚举从中间截断的情况即可，注意边界
for(int z=l;z<r;z++)
dp[l][r]=max(dp[l][r],dp[l][z]+dp[z+1][r]);
}

}

cout<<dp[0][str.size()-1]<<endl;

}

}