POJ 2955 brackets

Brackets
Time Limit: 1000MS Memory Limit: 65536K
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

Source

Stanford Local 2004

题意：给定只包含‘（’ ‘）'[' ']'字符串 , 求正确匹配的最长长度。

思路：区间DP + 记忆化搜索

区间DP：分治思想。 这里用记忆化搜索实现。

ans[I][J] 意为从i到j字符串的最长匹配长度

转移方程 从start到finsh 中若能正确匹配ch[start] == ch[k]

则ans[start][finsh] = max( ans[start][finsh] , dp(start+1,k-1) + dp(k+1,finsh) +2) ;

意为将start 到 finsh以k为中心分成两半 ，start+1 to k-1 和 k+1 to finsh 。

若不匹配 则 ans[start][finsh] = max( ans[start][finsh] , ans[start+1][finsh]) ;

意为，从已经匹配的start+1 to finsh 的结果 与 当前保存的start to finsh 结果取较大值。

从这也可看出搜索时割分时入栈的顺序 ： 1 ：begin to end 2： begin+1 to end 。。start to end 。。end-1 to end

ps brackets ： 方括号

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std ;
const int maxn = 100+10 ;
int brackets[maxn][maxn] ;
char ch[maxn] ;
bool judge(int start , int finsh) {
if( ch[start]=='('&&ch[finsh]==')') return true ;
if( ch[start]=='['&&ch[finsh]==']') return true ;
return false ;
}
int dp(int beginn , int endd  ) {
if( beginn > endd ) return 0 ;
int& ans = brackets[beginn][endd] ;
if(ans > 0 ) return ans ;
dp(beginn+1,endd) ;
for( int i = beginn+1 ; i <= endd ; ++i ) {
if( judge(beginn,i)  ) {
ans = max(ans, dp(beginn+1, i-1) + dp(i+1,endd)+2) ;
}
else
ans = max(ans, brackets[beginn+1][endd]) ;
}
return  ans ;
}
int main( ) {
while(~scanf("%s",ch)) {
if(ch[0] == 'e') break;
memset(brackets,0,sizeof(brackets)) ;
dp(0,strlen(ch)-1) ;
cout <<brackets[0][strlen(ch)-1]<< endl ;
}
return 0 ;
}