POJ -- 2955

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 … a_n, 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, …, i_m where 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im 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：动态规划，设dp[i][j]表示从i到j这段子串匹配的最大长度，则状态转移方程分两种情况，1.若从i到j-1这些字符中没有一个能与j匹配，则 dp[i][j] = dp[i][j-1]，这是显然的;2.若从i到j-1中有字符能与j匹配（可能不止一个，并设他们组成集合为A），则 dp[i][j] = max(dp[i][j],dp[i][k-1]+dp[k+1][j-1]+2)（k属于集合A）,加2是因为一旦匹配成功一次长度就会增加2.

#include<cstdio>
#include<cstring>
#include<iostream>
#define MAXN 111
using namespace std;
int dp[MAXN][MAXN];
char str[MAXN];
bool OK(int i,int j){return (str[i] == '(' && str[j] ==')') || (str[i] == '[' && str[j] == ']');}
int main(){
    //freopen("in.cpp","r",stdin);
    while(~scanf("%s",str) && strcmp(str,"end")){
        memset(dp,0,sizeof(dp));
        int len = strlen(str);
        for(int i = 1;i < len;i ++){
            for(int j = i-1;j >= 0;j--){
                dp[j][i] = dp[j][i-1];
                for(int k = j;k < i;k ++)
                    if(OK(k,i)) dp[j][i] = max(dp[j][i],dp[j][k-1] + dp[k+1][i-1]+2);
            }
        }
        printf("%d
",dp[0][len-1]);
        memset(str,0,sizeof(str));
    }
    return 0;
}