【UVA】11437Triangle Fun（简单几何）

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先求出在A,B,C上的三等分点在，这里使用向量运算进行加减就行了。

之后通过求出的三等分点和顶点的连线，求出3个交点。

最后用求出的三个交点算出面积。

注意：由于可能是钝角三角形，需要求其绝对值。

1411642811437Triangle FunAcceptedC++0.0152014-08-30 03:27:36

#include <iostream>#include <cstdlib>#include <cstdio>#include <string>#include <cstring>#include <cmath>#include <vector>#include <queue>#include <stack>#include <algorithm>using namespace std;const double eps = 1e-10;struct Point{double x, y;Point(double x = 0, double y = 0) : x(x), y(y) { }bool operator < (const Point& a) const{if(a.x != x) return x < a.x;return y < a.y;}};typedef Point Vector;Vector operator + (Vector A, Vector B) { return Vector(A.x+B.x, A.y+B.y); }Vector operator - (Point A, Point B)   { return Vector(A.x-B.x, A.y-B.y); }Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }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(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }double Length(Vector A) { return sqrt(Dot(A, A)); }double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }double Cross(Vector A, 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); }Vector Rotate(Vector A, double rad){return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));}Point GetIntersection(Point P, Vector v, Point Q, Vector w){Vector u = P-Q;double t = Cross(w, u) / Cross(v, w);return P+v*t;}bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2){double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1);double c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1);return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;}double PolygonArea(Point* p, int n){double area = 0;for(int i = 1; i < n-1; i++)area += Cross(p[i]-p[0], p[i+1]-p[0]);return area;}bool OnSegment(Point p, Point a1, Point a2){return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;}#define MAXD 50int main(){    int T;    Point P[MAXD];    scanf("%d",&T);    while(T--){        for(int i =  1 ; i <= 3 ; i++)            scanf("%lf%lf",&P[i].x,&P[i].y);        P[4].x = (2 * P[2].x + P[3].x) / 3;   //D        P[4].y = (2 * P[2].y + P[3].y) / 3;        P[5].x = (2 * P[3].x + P[1].x) / 3;   //E        P[5].y = (2 * P[3].y + P[1].y) / 3;        P[6].x = (2 * P[1].x + P[2].x) / 3;   //F        P[6].y = (2 * P[1].y + P[2].y) / 3;        Point a = GetIntersection(P[4],P[1] - P[4],P[5],P[2] - P[5]);        //cout << a.x << " " << a.y << endl;        Point b = GetIntersection(P[4],P[1] - P[4],P[6],P[3] - P[6]);        //cout << b.x << " " << b.y << endl;        Point c = GetIntersection(P[5],P[2] - P[5],P[6],P[3] - P[6]);        //cout << c.x << " " << c.y << endl;        double ans = Area2(a,b,c) / 2;        ans = floor(ans + 0.5);        printf("%.f\n",fabs(ans));    }    return 0;}

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