Intersecting Lines

Posted 2021-01-29 jeffrey-y

总时间限制: 

1000ms

 

内存限制: 

65536kB

描述

We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three ways: 1) no intersection because they are parallel, 2) intersect in a line because they are on top of one another (i.e. they are the same line), 3) intersect in a point. In this problem you will use your algebraic knowledge to create a program that determines how and where two lines intersect. 
Your program will repeatedly read in four points that define two lines in the x-y plane and determine how and where the lines intersect. All numbers required by this problem will be reasonable, say between -1000 and 1000.

输入

The first line contains an integer N between 1 and 10 describing how many pairs of lines are represented. The next N lines will each contain eight integers. These integers represent the coordinates of four points on the plane in the order x1y1x2y2x3y3x4y4. Thus each of these input lines represents two lines on the plane: the line through (x1,y1) and (x2,y2) and the line through (x3,y3) and (x4,y4). The point (x1,y1) is always distinct from (x2,y2). Likewise with (x3,y3) and (x4,y4).

输出

There should be N+2 lines of output. The first line of output should read INTERSECTING LINES OUTPUT. There will then be one line of output for each pair of planar lines represented by a line of input, describing how the lines intersect: none, line, or point. If the intersection is a point then your program should output the x and y coordinates of the point, correct to two decimal places. The final line of output should read "END OF OUTPUT".

【题解】

计算几何模板题。关键是把和interset()函数相关的计算模板背下。

【代码】

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cmath>
 4 using namespace std;
 5 #define CPoint CVector
 6 #define PI acos(-1)
 7 #define INF 1e20
 8 #define EPS 1e-12
 9 
10 struct CVector
11 {
12     double x, y;
13     CVector(double _x, double _y) :
14         x(_x), y(_y) {}
15     CVector() :x(INF), y(INF) {}
16 };
17 
18 struct CLine
19 {
20     CPoint a, b;
21     CLine(CPoint _a, CPoint _b) {
22         a = _a;
23         b = _b;
24     }
25 };
26 
27 bool isZero(double x) {
28     return -EPS < x && x < EPS;
29 }
30 
31 CVector operator + (CVector p, CVector q) {
32     return CVector(p.x + q.x, p.y + q.y);
33 }
34 
35 CVector operator - (CVector p, CVector q){
36     return CVector(p.x - q.x, p.y - q.y);
37 }
38 
39 double operator ^(CVector p, CVector q) {
40     return p.x * q.y - p.y * q.x;
41 }
42 
43 double area(CVector p, CVector q) {
44     return (p ^ q) / 2;
45 }
46 
47 double operator *(CVector p, CVector q) {
48     return p.x*q.x + p.y*q.y;
49 }
50 
51 CVector operator *(double k, CVector p) {
52     return CVector(k*p.x, k*p.y);
53 }
54 
55 double length(CVector p) {
56     return sqrt(p * p);
57 }
58 
59 double dist(CPoint p, CLine l) {
60     return fabs((p - l.a) ^ (l.b - l.a)) / length(l.b - l.a);
61 }
62 
63 CPoint interset(CLine l, CLine m, string &msg) {
64     double x = area(m.a - l.a, l.b - l.a);
65     double y = area(l.b - l.a, m.b - l.a);
66     if (isZero(x + y)) {
67         if (isZero(dist(l.a, m)))
68             msg = "LINE";
69         else msg = "NONE";
70         return CPoint();
71     }
72     msg = "POINT";
73     return m.a + (x / (x + y)) * (m.b - m.a);
74 }
75 
76 int main()
77 {
78     int N = 0;
79     cin >> N;
80     printf("INTERSECTING LINES OUTPUT
");
81     while (N--) {
82         int x1, y1, x2, y2, x3, y3, x4, y4;
83         string msg;
84         cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3 >> x4 >> y4;
85         CLine l = CLine(CPoint(x1, y1), CPoint(x2, y2));
86         CLine m = CLine(CPoint(x3, y3), CPoint(x4, y4));
87         CPoint p = interset(l, m, msg);
88         if (p.x == INF) printf("%s
", msg.c_str());
89         else printf("%s %.2f %.2f
", msg.c_str(), p.x, p.y);
90     }
91     printf("END OF OUTPUT
");
92     //system("pause");
93     return 0;
94 }