如何找到通过给定点（a，b）和（x，0）的线的最大y轴截距？

Question

I have given points (a, b) and then I have given points (x, 0) . Now for each point (x, 0) I make lines with all points (a, b) (see image).
For each point (x, 0) I have to return indexes of thease point/points (a, b) which y-intercept of line through this point/points and point (x, 0) is largest.
All x values of points (x, 0) are greater than maximum a . Numbers a, b, x are positive integers.

Example:

Input

3 4 (number of (a, b) points and number of (x, 0) points - let's call them m and n)
5 3 (point A, index 0)
14 1 (point C, index 1)
10 2 (point B, index 2)
16 20 40 15 (x values of points (x, 0)) 

Output

1
0 2
0
1

My solution:

int main() {
    int m, n;
    cin >> m >> n;
    vector<pair<int, int>> pointsAB(m);

    for (int i = 0; i < m; ++i) {
        cin >> pointsAB[i].first >> pointsAB[i].second;
    }

    for (int j = 0; j < n; ++j) {
        int currX;
        double minSlope = 1.00;
        vector<int> indexes;
        cin >> currX;
        for (int i = 0; i < m; ++i) {
            int a = pointsAB[i].first, b = pointsAB[i].second;
            double currSlope = -((double)b) / (currX - a);
            if (currSlope < minSlope) {
                indexes.clear();
                minSlope = currSlope;
                indexes.push_back(i);
            }
            else if (currSlope == minSlope) {
                indexes.push_back(i);
            }
        }

        cout << indexes[0];
        for (int k = 1; k < indexes.size(); ++k) {
            cout << " " << indexes[k];
        }
        cout << '\n';
    }

    return 0;
}

My solution to this problem has time complexity O(m * n) but this doesn't seem very efficient to me. My question is can be this problem solved with better time complexity and how?

Answer 1

Build convex hull for a/b points, get only top half (really you need only right leg of upper envelope) in order from the most right point

Sort x-points

Complexity is about O(mlogm + nlogn ) (depending on hull and sorting methods)

Walk through x-list in order from small values, finding the best points of a/b set. Note that this process is linear O(n+m) (we will find next best a/b point only to the left of current one - imagine rotating stick, one end moves along OX axis, another end rests on the a/b point set)

Answer 2

Most of the steps here seem fairly obvious:

1. read in the points
2. read in the X-intercept for each line (did I just invent "x-intercept"?)
3. compute the slopes of the lines
4. select the smallest slope
5. find all the lines with that slope
6. print out the results

I believe all of those can be done with O(N) complexity, so it should be O(N) overall.

#include <vector>
#include <algorithm>
#include <iostream>
#include <iterator>

struct point {
    int x;
    int y;

    friend std::istream &operator>>(std::istream &is, point &p) {
        return is >> p.x >> p.y;
    }

    friend std::ostream &operator<<(std::ostream &os, point const &p) {
        return os << "(" << p.x << ", " << p.y << ")";
    }
};

struct slope_index {
    double slope;
    int index;

    bool operator<(slope_index const &other) const {
        return slope < other.slope;
    }

    bool operator==(slope_index const &other) const {
        return slope == other.slope;
    }
};

int main() {
    int N;
    std::cin >> N;

    // read in the points
    std::vector<point> points;
    std::copy_n(std::istream_iterator<point>(std::cin), N, std::back_inserter(points));

    // read in the X-intercept for each point:
    std::vector<int> Xs;
    std::copy_n(std::istream_iterator<int>(std::cin), N, std::back_inserter(Xs));

    // compute the slopes
    std::vector<slope_index> slopes;
    int i = 0;
    std::transform(points.begin(), points.end(),
        Xs.begin(),
        std::back_inserter(slopes),
        [&](point const &p, int currX) { return slope_index{ p.y / double(p.x - currX), i++ }; });

    // find the smallest slope
    auto v = *std::min_element(slopes.begin(), slopes.end());

    // find all the lines with that slope:
    auto pos = std::partition(slopes.begin(), slopes.end(), [&](auto const &s) { return v == s; });

    // print out the results:
    for (auto s = slopes.begin(); s != pos; ++s)
        std::cout << points[s->index];
}