Learning algorithm: Computational Geometry foundation

【definition】

[] Polar angle polar coordinate system phi, it can be directly regarded as the straight line slope, since the angle of the slope may be expressed as

Computational Geometry is the use of computer code to simulate the real value and the geometric calculation process, so you need to create the corresponding elements, and the most basic element is the point, and the point position can be expressed in two-dimensional space

In the same two-dimensional space, storage methods and vectors point similar (but actually very different meaning)

struct V
{
    double x, y;
    double the;
    double angle() {// polar angle is obtained
        return atan2(y, x);
    }
    V(double X = 0, double Y = 0) { //初始化 x = X, y = Y; ang = atan2(y, x); } bool operator ==(const V &b) { return cmp(x - b.x) && cmp(y - b.y); } };

By renaming make some points and vectors coincide on some usage

typedef V  P;

Defined operation (calculation points and vectors of the individual have the coincident)

V operator +(V a, V b) { return V(a.x + b.x, a.y + b.y); }
V operator -(V a, V b) { return V(a.x - b.x, a.y - b.y); }
V operator *(V a, double b) { return V(a.x *b, a.y*b); }
V operator /(V a, double b) { return V(a.x / b, a.y / b); } // cross product
double cross(V a, V b)
{
    return a.x*b.y - a.y*b.x; } //点积 double dot(V a, V b) { return a.x*b.x + a.y*b.y; }

According to the knowledge of geometry in high school, we know two points can be represented by a line segment

So there are segments defined

struct L
{
    P s, t;
    double the;
    L(P X = V(), P Y = V())
    {
        s = X, t = Y, ang = (Y - X) .angle ();
    }
};

Through their respective operations are a variety of requirements intersection, parallel, cross-product, dot product obtaining complete

(Proved relatively simple)

By comparing the slope of the line size //
bool operator <(const L &a, const L &b)
{
    double r = reasonable.The - B;
    if (cmp(r) != 0)    return cmp(r) == -1;
    // same polar angle, the greater the default partial
    return cmp(cross(a.t - a.s, b.t - a.s)) == -1;
}
 Analyzing line parallel //
bool is_parallel(L a, L b)
{
    return cmp(cross(a.t - a.s, b.t - b.s)) == 0; } //查找交点 P intersection(L a, L b) { return a.s + (a.t - a.s)*(cross(b.t - b.s, a.s - b.s)) / cross(a.t - a.s, b.t - b.s); }

Determined surface may have a plurality of segments or a plurality of points determined

Polygon area [seeking]

Is the cross product can be represented by: two line segments into a parallelogram area

Then one by one is obtained by the polygon edge and an area surrounded origin

Because the cross product of both positive and negative,

This principle can be obtained by polygon area

double area(P *p, int n)
{
    double res = 0;
    p[n + 1] = p[1];
    for (int i = 1; i <= n; i++)
        res += cross(p[i], p[i + 1]);
    return fabs(res / 2); }

There are a variety of operations between these elements is determined and

Point is in the linear side

bool is_right(L a, P b)
// determines whether the point b (and even into the starting point and the linear vector direction opposite to the vector direction) on the right side of a straight line
{
    return cmp(cross(a.t - a.s, b - a.s)) < 0;
}

Seeking whether the points within the polygon

By calculating the number of sides is equal on both sides of the point, if they are equal then the points within the polygon

Meanwhile, the point is calculated on a straight line, find the point where the straight line

Point is within the polygon

Seeking two vectors intersect

Two line segments intersect

Through the above two knowledge points:

Possible to find two polygons intersect

Two polygons intersect

Learning algorithm: Computational Geometry foundation

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