POJ 2955 Brackets

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2938		Accepted: 1516

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

Source

Stanford Local 2004

 

解题：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <climits>
#include <algorithm>
#include <cmath>
#define LL long long
using namespace std;
const int maxn = 102;
char str[maxn<<1];
int dp[maxn][maxn];
int main() {
    int i,j,k,len,t,ans;
    while(gets(str)&&strcmp(str,"end")) {
        len = strlen(str);
        memset(dp,0,sizeof(dp));
        ans = 0;
        for(i = 0; i < len; i++) {
            for(j = 0,k = i; k < len; k++,j++) {
                if((str[j] == '(' && str[k] == ')') || (str[j] == '[' && str[k] == ']')) {
                    dp[j][k] = dp[j+1][k-1] + 2;
                }
                for(t = j+1; t < k; t++)
                    if(dp[j][t]+dp[t][k] > dp[j][k]) dp[j][k] = dp[j][t]+dp[t][k];
                if(dp[j][k] > ans) ans = dp[j][k];
            }

        }
        printf("%d
",ans);
    }
    return 0;
}