POJ2826 An Easy Problem?!（线段交点，三角形面积）

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <cmath>
 6 using namespace std;
 7 
 8 const double EPS = 1e-10;    //精度系数
 9 
10 struct Point {
11     double x, y;
12     Point(double x = 0, double y = 0) :x(x), y(y) {}
13 };    //点的定义
14 
15 typedef Point Vector;    //向量的定义
16 
17 int dcmp(double x) {
18     if (fabs(x) < EPS)return 0; else return x < 0 ? -1 : 1;
19 }    //与0的关系
20 
21 Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }    //向量加法
22 Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }    //向量减法
23 Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }        //向量数乘
24 bool operator == (const Point& a, const Point& b) {
25     return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
26 }    //点相等
27 
28 double Angle(Vector A) { return atan2(A.y, A.x); }    //向量极角
29 
30 double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }    //向量叉乘
31 
32 bool SegmentProperIntersection(Point& a1, Point& a2, Point& b1, Point& b2) {
33     double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),
34         c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
35     return dcmp(c1)*dcmp(c2) <= 0 && dcmp(c3)*dcmp(c4) <= 0;
36 }    //判断两线段相交（含端点）
37 
38 Point LineIntersectionPoint(Point A1, Point B1, Point A2, Point B2) {
39     double t1 = ((A1.y - A2.y)*(B2.x - A2.x) - (A1.x - A2.x)*(B2.y - A2.y)) /
40         ((B1.x - A1.x)*(B2.y - A2.y) - (B1.y - A1.y)*(B2.x - A2.x));
41     return A1 + (B1 - A1)*t1;
42 }    //返回直线交点
43 
44 double Area2(Point& A, Point& B, Point& C) { return Cross(B - A, C - A); }
45 //AB,AC向量围成的三角形面积的两倍
46 
47 Point A1, B1, A2, B2;
48 
49 int main() {
50     int t;
51     cin >> t;
52     while (t--) {
53         scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &A1.x, &A1.y, &B1.x, &B1.y, &A2.x, &A2.y, &B2.x, &B2.y);
54         Vector v1 = B1 - A1, v2 = B2 - A2, u(1, 0);
55         if (dcmp(Cross(v1, u)) == 0 || dcmp(Cross(v1, u)) == 0) { printf("0.00\n"); continue; }    //情况1
56         if (!SegmentProperIntersection(A1, B1, A2, B2)) { printf("0.00\n"); continue; }    //情况2
57         double d1 = Angle(v1), d2 = Angle(v2);
58         if (dcmp(d1) < 0) { swap(A1, B1); v1 = v1*-1; d1 = Angle(v1); }
59         if (dcmp(d2) < 0) { swap(A2, B2); v2 = v2*-1; d2 = Angle(v2); }
60         if (dcmp(d1 - d2) == 0) { printf("0.00\n"); continue; }    //情况3
61         if (d1 > d2) { swap(d1, d2); swap(v1, v2); swap(A1, A2); swap(B1, B2); }
62         if (B2.x >= B1.x || B1 == B2) { printf("0.00\n"); continue; }    //情况4 5
63         Point C1 = LineIntersectionPoint(A1, B1, A2, B2), tmp, N1, N2;    //线段交点
64         if (B1.y < B2.y)tmp = B1, N1 = A2, N2 = B2;
65         else tmp = B2, N1 = A1, N2 = B1;    //求出B1 B2 中y值较小的点
66         Point C2 = LineIntersectionPoint(tmp, tmp + u, N1, N2);    //三角形第三个点
67         double ans = fabs(Area2(C1, C2, tmp)) / 2;
68         printf("%.2lf\n", ans);
69     }
70 }