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POJ 2955 Brackets (区间dp 括号匹配)

2024-10-07 12:57:01   211人阅读


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, and
  • if a and b are regular brackets sequences, thenab 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 ofs. That is, you wish to find the largest m such that for indicesi1, 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

题目链接：

id=2955">http://poj.org/problem?

id=2955

题目大意：给一个括号序列，问序列中合法的括号最多有多少个。若A合法。则[A]，(A)均合法，若A,B合法则AB也合法

题目分析：和POJ 1141那道经典括号匹配类似，这题更简单一些。想办法把问题转化，既然要求最大的括号匹配数，我们考虑加最少的括号。使得整个序列合法，这样就转变成1141那题。开下脑动类比二分图最大匹配的性质，最大匹配＋最大独立集＝点数，显然要增加最少的点使序列合法，则加的最少的点数即为|最大独立集|。我们要求的是原序列的|最大匹配|，以上纯属yy，以下给出转移方程，和1141一模一样
dp[i][i] = 1;
然后枚举区间长度
1）外围匹配：dp[i][j] = dp[i + 1][j - 1];
2）外围不匹配。枚举切割点：dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]); (i <= k < j)

#include <cstdio>
#include <cstring>
#include <algorithm>
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]);
    }
}


POJ 2955 Brackets (区间dp 括号匹配)


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