Triangle Fun UVA - 11437

Triangle Fun

UVA - 11437

题意：给三角形，求一些三等分点，线段交点，求面积。

简单题~

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int inf = 0x3f3f3f3f;
 4 const double eps = 1e-12;
 5 const double pi = acos(-1.0);
 6
 7 struct Point {
 8     double x,y;
 9     Point (double x = 0, double y = 0) : x(x), y(y) {}
10 };
11 typedef Point Vector;
12 Vector operator + (Vector a, Vector b) {
13     return Vector (a.x + b.x, a.y + b.y);
14 }
15 Vector operator * (Vector a, double s) {
16     return Vector (a.x * s, a.y * s);
17 }
18 Vector operator / (Vector a, double p) {
19     return Vector (a.x / p, a.y / p);
20 }
21 Vector operator - (Point a, Point b) {
22     return Vector (a.x - b.x, a.y - b.y);
23 }
24 bool operator < (Point a, Point b) {
25     return a.x < b.x || (a.x == b.x && a.y < b.y);
26 }
27 int dcmp (double x) {
28     if(fabs(x) < eps) return 0;
29     return x < 0 ? -1 : 1;
30 }
31 bool operator == (const Point &a, const Point &b) {
32     return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
33 }
34 double Angel (Vector a) {
35     return atan2(a.y, a.x);
36 }
37 double Dot(Vector a, Vector b) {
38     return a.x * b.x + a.y * b.y;
39 }
40 double Length (Vector a) {
41     return sqrt(Dot(a, a));
42 }
43 double Angle (Vector a, Vector b) {
44     return acos(Dot(a, b) / Length(a) / Length(b));
45 }
46 double Cross (Vector a, Vector b) {
47     return a.x * b.y - a.y * b.x;
48 }
49 //两直线交点
50 Point GetLineIntersection (Point p, Vector v, Point q, Vector w) {
51     Vector u = p - q;
52     double t1 = Cross(w, u) / Cross(v, w);
53     double t2 = Cross(v, u) / Cross(v, w);
54     return p + v * t1;  //  return q + w * t2;
55 }
56 //多边形面积
57 double ConvexPolygonArea(Point *p, int n) {
58     double area = 0;
59     for(int i = 1; i < n-1; i++) {
60         area += Cross(p[i] - p[0], p[i+1] - p[0]);
61     }
62     return area / 2;
63 }
64
65 Point p[3],ch[3];
66 int main(){
67     int t;
68     //freopen("in.txt", "r", stdin);
69     scanf("%d", &t);
70     while(t--){
71         for(int i = 0; i < 3; i++){
72             scanf("%lf %lf", &p[i].x, &p[i].y);
73         }
74         Point d = p[1] + (p[2] - p[1])/3;
75         Point e = p[2] + (p[0] - p[2])/3;
76         Point f = p[0] + (p[1] - p[0])/3;
77         ch[0] = GetLineIntersection(p[0], d-p[0], p[1], e-p[1]);
78         ch[1] = GetLineIntersection(p[2], f-p[2], p[1], e-p[1]);
79         ch[2] = GetLineIntersection(p[2], f-p[2], p[0], d-p[0]);
80         double area = ConvexPolygonArea(ch,3);
81         printf("%.0lf\n", area);  //这里是四舍五入，如果强制转换是截断，会WA
82     }
83 }