计算三角形内的晶格点 [英] Counting lattice points inside a triangle

问题描述

我已经指出了一个大三角形，所以将其称为a，b，c。 （a =（x，y）依此类推。）

I've points for a big triangle lets call it a, b, c. (a = (x, y) and so on).

现在，我想计算该三角形围成的区域内的积分点数，因此我首先看一下在匹克定理中。我考虑的第二种方法是生成一个以三角形的最大值，最小值为边界的点的列表，然后检查每个点是否位于三角形内。

Now I want to count the number of integral points inside the area enclosed by this triangle, so I first looked at Pick's theorem. The second approach that I considered was generating a list of points bounded by max, min of the triangle, and then checking if each point lies inside the triangle.

我使用了重心坐标法可以做到这一点。它可以工作，但是我的三角形很大，我的程序基本上是跨点的蛮力。我如何改进此算法？

I used the Barycentric coordinates method to do this. It works however my triangle is pretty huge and my program is basically a brute force across points. How do I improve on this algorithm?

可以在这里找到我的代码： https://bpaste.net/show/58433b6e389c

My code can be found here: https://bpaste.net/show/58433b6e389c

推荐答案

可以并且应该解决此问题使用皮克定理。阅读本文可全面了解它的工作原理，它将适用于您能想到的任何多边形。因此， Pick说说，如果要计算多边形的面积，则公式为 area = noOfInsidePoints + noOfBoundaryPoints / 2-1 。要计算任何多边形的面积，可以使用以下代码，其中 pc 是表示多边形顶点的结构数组。

This problem can and should be solved using Pick's theorem. Read the article to have a full understanding on how it works and it will work for any polygon you can think of. So, "Pick says" that if you want to compute the area of a polygon you have the formula area = noOfInsidePoints + noOfBoundaryPoints /2 - 1. To compute the area of any polygon you can use the following code, where pc is an array of structures representing the vertices of the polygon.

float computeArea()
{
    float area = 0;
    for(int i=1;i<=n;++i) // n is the total number of vertices
        area += (pc[i].x*pc[i+1].y - pc[i+1].x*pc[i].y );
    if(area < 0)
        area *= (-1);
}

计算出面积后，我们必须计算所有多边形边缘。可以使用以下方法完成此操作：

Having the area computed, we have to count the number of point from all of the polygon edges. This can be done using the following:

int getBoundaryPoints()
{
    long left=0, right=0;
    int noPoints = 0;
    for(int i=1;i<=n;++i)
    {
        st = abst( pc[i].x - pc[i+1].x );
        right = abs( pc[i].y - pc[i+1].y );
        if(right == 0)
            right = left;
        if(left == 0)
            left = right;
        noPoints += gcd(left, right) +1;
    }
}

也对此进行了计算，我们可以找到

Having this computed too, we can find the number of point inside

noPointsInside = (computeArea() - (getBoundaryPoints() - n)) / 2 + 1;

最终时间复杂度：O（N）
最终内存复杂度：O（N）

Final time complexity: O(N) Final memory complexity: O(N)

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