algorithm – 用于“查找包含P中最大点数的行”的哈希函数

2023年4月6日 238次阅读

以下是“编程访谈元素”一书中的一段：

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.

在解决方案部分中说：

 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.

这是Line的哈希函数：

// 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);
}

这里是斜坡和拦截的声明：

 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xor仍然是唯一的？

最佳答案 我们可以尝试以下简单的算法：

>使用键创建哈希映射作为线的一个(斜率,截距)对,并将值作为具有相同斜率截距的线的数量.
>对于所有点对(O(n ^ 2)对)计算(斜率,截距)值并增加hashmap中的相应键(在最坏的情况下,它将消耗O(n ^ 2)内存但是如果有是很多共线点,那么平均空间复杂度应该很低).
>输出线,即散列图中具有最高计数的(斜率,截距)(为此,您需要遍历hasmap,在最坏的情况下将包含O(n ^ 2)个条目).

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