UVa 11437 Triangle Fun(几何)

最新推荐文章于 2022-10-22 13:25:38 发布

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本文提供UVa11437 TriangleFun问题的解题思路及代码实现，采用向量计算解决几何问题，包括点、向量定义、向量运算、点到直线的距离等关键函数。

题目链接：UVa 11437 Triangle Fun

几何，用向量可以做。

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

const double eps = 1e-10;

struct Point
{
	double x, y;
	Point(double x=0, double y=0):x(x),y(y) { }
};

typedef Point Vector;

Vector operator + (const Vector& A, const Vector& B)
{
    return Vector(A.x+B.x, A.y+B.y);
}

Vector operator - (const Point& A, const Point& B)
{
    return Vector(A.x-B.x, A.y-B.y);
}

Vector operator * (const Vector& A, double p)
{
    return Vector(A.x*p, A.y*p);
}

bool operator < (const Point& a, const Point& b)
{
	return a.x < b.x || (a.x == b.x && a.y < b.y);
}

int dcmp(double x)
{
    if(fabs(x) < eps)
        return 0;
    else
        return x < 0 ? -1 : 1;
}

bool operator == (const Point& a, const Point &b)
{
	return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
}

double Dot(const Vector& A, const Vector& B)
{
    return A.x*B.x + A.y*B.y;
}

double Length(const Vector& A)
{
    return sqrt(Dot(A, A));
}

double Angle(const Vector& A, const Vector& B)
{
    return acos(Dot(A, B) / Length(A) / Length(B));
}

double Cross(const Vector& A, const Vector& B)
{
    return A.x*B.y - A.y*B.x;
}

double Area2(Point A, Point B, Point C)
{
    return Cross(B - A, C - A);
}

bool SegmentProperIntersection(const Point& a1, const Point& a2, const Point& b1, const Point& b2)
{
	double c1 = Cross(a2-a1,b1-a1), c2 = Cross(a2-a1,b2-a1),
		c3 = Cross(b2-b1,a1-b1), c4=Cross(b2-b1,a2-b1);
	return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
}

bool OnSegment(const Point& p, const Point& a1, const Point& a2)
{
	return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;
}

//求两直线交点，先确保两直线有唯一交点，当且仅当Cross(v, w)非0
Point GetLineIntersection(const Point& P, const Point& v, const Point& Q, const Point& w)
{
	Vector u = P-Q;
	double t = Cross(w, u) / Cross(v, w);
	return P+v*t;
}

Vector Rotate(const Vector& A, double rad)
{
	return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));
}
double DistanceToLine(Point P, Point A, Point B)
{
    Vector v1 = B - A, v2 = P - A;
    return fabs(Cross(v1, v2)) / Length(v1); // 如果不取绝对值，得到的是有向距离
}


int main()
{
    int T, n;
    Point a, b, c, d, f, e, r, p, q;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%lf%lf", &a.x, &a.y);
        scanf("%lf%lf", &b.x, &b.y);
        scanf("%lf%lf", &c.x, &c.y);

        d = b + (c - b) * (1.0 / 3.0);
        f = a + (b - a) * (1.0 / 3.0);
        e = c + (a - c) * (1.0 / 3.0);

        r = GetLineIntersection(a, a - d, c, c - f);
        p = GetLineIntersection(a, a - d, b, b - e);
        q = GetLineIntersection(b, b - e, c, c - f);

        int res = (int)(fabs(Area2(r, p, q) * 0.5) + 0.5);
        printf("%d\n", res);
    }
    return 0;
}