POJ 2955 Brackets (區間dp 括號匹配)

POJ 2955 Brackets (區間dp 括號匹配)

 

Brackets Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3951   Accepted: 2078

 

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, andif 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

題目鏈接：http://poj.org/problem?id=2955

題目大意：給一個括號序列，問序列中合法的括號最多有多少個，若A合法，則[A]，(A)均合法，若A,B合法則AB也合法

題目分析：和POJ 1141那道經典括號匹配類似，這題更簡單一些，想辦法把問題轉化，既然要求最大的括號匹配數，我們考慮加最少的括號，使得整個序列合法，這樣就轉變成1141那題，開下腦動類比二分圖最大匹配的性質，最大匹配＋最大獨立集＝點數，顯然要加入最少的點使序列合法，則加的最少的點數即為|最大獨立集|，我們要求的是原序列的|最大匹配|，以上純屬yy，下面給出轉移方程，和1141一模一樣
dp[i][i] = 1;
然後枚舉區間長度
1）外圍匹配：dp[i][j] = dp[i + 1][j - 1];
2）外圍不匹配，枚舉分割點：dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]); (i <= k < j)

#include 
#include 
#include 
using namespace std;
int const INF = 0x3fffffff;
char s[205];
int dp[205][205];

int main()
{
    while(scanf("%s", s) != EOF && strcmp(s, "end") != 0)
    {
        int len = strlen(s);
        memset(dp, 0, sizeof(dp));
        for(int i = 0; i < len; i++)
            dp[i][i] = 1;
        for(int l = 1; l < len; l++)
        {
            for(int i = 0; i < len - l; i++)
            {
                int j = i + l;
                dp[i][j] = INF;
                if((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']'))
                    dp[i][j] = dp[i + 1][j - 1];
                for(int k = i; k < j; k++)
                    dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]);
            }
        }
        printf("%d\n", len - dp[0][len - 1]);
    }
}