algorithm - "find the line that contains the maximum number of points in P"的哈希函数

Question

这是编程面试基础一书中的一段话:

Let P be a set of n points in the plane. Each point has integer coordinates. Design an efficient algorithm for computing a line that contains the maximum number of points in P.

解决方案部分是这样说的:

 Every pair of distinct points defines a line. We can use a hash table
H to map lines to the set of points in P that lie on them.

下面是行的散列函数:

// Hash function for Line.
struct HashLine {
   size_t operator()(const Line& l) const {
       return hash <int >()(l.slope.first) ^ hash <int >()(l.slope.second) ^ hash <int >()(l.intercept.first) ^ hash <int >()(l.intercept.second);
}

这里是斜率和截距的声明:

 pair <int, int> get_canonical_fractional(int a, int b) {
    int gcd = GCD(abs(a), abs(b));
    a /= gcd, b /= gcd;
    return b < 0 ? make_pair(-a, -b) : make_pair(a, b);
 }

     // Line function of two points , a and b, and the equation is
     // y = x(b.y - a.y) / (b.x - a.x) + (b.x * a.y - a.x * b.y) / (b.x - a.x).
     struct Line {
     Line(const Point& a, const Point& b)
     : slope(a.x != b.x ? get_canonical_fractional(b.y - a.y, b.x - a.x) : make_pair(1, 0))
     , intercept(a.x != b.x ? get_canonical_fractional(b.x * a.y - a.x * b.y,  b.x - a.x) : make_pair(a.x, 1))
     {}

       ...
     // Store the numerator and denominator pair of slope unless the line is
     // parallel to y-axis that we store 1/0.  
     pair <int, int> slope;
     // Store the numerator and denominator pair of the y-intercept unless
     // the line is parallel to y-axis that we store the x-intercept.     
     pair <int, int> intercept;
};

但是我们怎么知道如果斜率和截距对是唯一的，那么它们的异或仍然是唯一的？

Answer 1

我们可以试试下面的简单算法:

1. 创建一个散列映射，键为一条线的(slope, intercept)对，值为具有相同斜率截距的线的数量。
2. 对于所有点对(O(n^2) 对)计算(slope, intercept) 值并递增 hashmap 中的相应键(在最坏的情况，它会消耗 O(n^2) 内存，但如果有很多共线点，那么平均空间复杂度应该很低。
3. 输出行，即 hashmap 中计数最高的 (slope, intercept)(为此，您需要遍历 hasmap，在最坏的情况下，该 hasmap 将包含 O( n^2) 个条目)。