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  1. /* PRESET CODE BEGIN - NEVER TOUCH CODE BELOW */
  2.  
  3. //請完成兩個運算子覆載,err_detection和find_intersection函式
  4. #include <iostream>
  5. #include <cmath>
  6. using namespace std;
  7. const  double  TOL = 1.e-13;
  8. //異常處理類別
  9. class Err{
  10.   private:
  11.     int kind;//可用來判斷異常的種類，如兩點重疊(目前設定的值為1)、沒有交點、兩線重疊
  12.   public:
  13.     Err(int i) : kind(i){}
  14.     //完成運算子覆載功能，透過kind的值將不同種類的異常狀態回傳
  15.     friend ostream& operator<<(ostream& out, const Err& foo);
  16. };
  17. //生成點座標的類別
  18. class  Point{
  19.   private:
  20.     double x, y;
  21.  
  22.   public:
  23.     Point(double a = 0. , double b = 0.) : x(a) , y(b){}
  24.     //將交於一點的該點根據題意輸出
  25.     friend ostream& operator<<(ostream& out , const Point& pt){
  26.         return  out << "(" << (abs(pt.x) < TOL ? 0. : pt.x) 
  27.                     << "," << (abs(pt.y) < TOL ? 0. : pt.y)
  28.                     << ")";
  29.     }
  30.  
  31.     friend istream& operator>> (istream& in , Point& pt){
  32.         return in >> pt.x >> pt.y;
  33.     }
  34.     //完成運算子覆載功能，透過該函式比較兩點座標是否相等，相等回傳true
  35.     friend bool operator==(const Point& a, const Point& b);
  36.     friend class Line;
  37. };
  38. //透過兩點(Point物件)構成一線
  39. class  Line{
  40.   private:
  41.     Point pt1, pt2;
  42.     double a, b, c;//直線方程式為ax+by=c
  43.  
  44.   public:
  45.     Line(const Point& m, const Point& n) : pt1(m), pt2(n){
  46.         if(m == n){//透過運算子覆載功能比較兩點座標
  47.             throw(Err(1));//若無法構成線(兩點重疊)的情況，擲出Err物件
  48.         }
  49.         if(pt1.x == pt2.x){
  50.             a = 1;
  51.             b = 0;
  52.             c = pt1.x;
  53.         }
  54.         else{
  55.         a = (pt2.y - pt1.y) / (pt2.x - pt1.x); 
  56.         b = -1;
  57.         c = a * pt1.x - pt1.y;
  58.         }
  59.     }
  60.  
  61.     friend Point intersection(const Line& line1 , const Line& line2);
  62.  
  63.     //待完成的函式，功能主要是回傳異常的種類("沒有交點"或"兩線重疊")，沒有異常回傳0
  64.     friend int err_detection(const Line& line1 , const Line& line2);
  65.     //待完成的函式，計算出兩線交於一點的點座標，可透過x,y儲存計算出的xy座標
  66.     friend void find_intersection(const Line& line1, const Line& line2, double *x, double *y);
  67. };
  68. //先透過err_detection()判斷line1, line2是否有異常，之後再透過find_intersection()找出交於一點的點座標
  69. Point intersection(const Line& line1, const Line& line2){
  70.     double point_x, point_y;//交點的點座標
  71.     int errlog = err_detection(line1, line2);//先找出是否有異常
  72.     if (errlog != 0) throw(Err(errlog));
  73.     else{
  74.         find_intersection(line1, line2, &point_x, &point_y);//計算出交於一點的點座標
  75.         return Point(point_x, point_y);
  76.     } 
  77. }
  78.  
  79. int main(){
  80.     Point pt1, pt2, pt3, pt4;
  81.     while(cin >> pt1 >> pt2 >> pt3 >> pt4){
  82.         try{
  83.             Line line1(pt1, pt2);
  84.             Line line2(pt3, pt4);
  85.             cout << intersection(line1, line2) << endl;
  86.         }
  87.         catch(const Err& err){
  88.             cout << err << endl;
  89.         }  
  90.     }
  91.     return 0;
  92. }
  93.  
  94. /* PRESET CODE END - NEVER TOUCH CODE ABOVE*/
  95. int err_detection(const Line& line1 , const Line& line2){
  96.     double m,n,r;
  97.     m=fabs(line1.a/line2.a);
  98.     n=fabs(line1.b/line2.b);
  99.     r=fabs(line1.c/line2.c);             
  100.     if(m==n&&n==r){
  101.         return 2; 
  102.         //重和
  103.     }
  104.     else if(m==n&&n!=r){
  105.         return 3;
  106.         //沒焦點
  107.     }
  108.     else{
  109.         return 0;
  110.     }
  111. }
  112. void find_intersection(const Line& line1, const Line& line2, double *x, double *y){
  113.     double det=line1.a*line2.b-line1.b*line2.a;
  114.     double ans1;
  115.     double ans2;
  116.     ans1=(line1.c*line2.b-line1.b*line2.c)/det;
  117.     ans2=(line1.a*line2.c-line1.c*line2.a)/det;
  118.     x=&ans1;
  119.     y=&ans2;
  120.     //對應點 line1.a line1.b line1.c line2.a line2.b line2.c
  121. }
  122. ostream& operator<<(ostream& out, const Err& foo){
  123.         if (foo.kind == 1) {
  124.             return out << "兩點重疊";
  125.         }
  126.         else if (foo.kind == 2) {
  127.             return out << "兩線重疊";
  128.         } 
  129.         else if (foo.kind == 3) {
  130.             return out << "沒有交點";
  131.         }
  132. }
  133. bool operator==(const Point& a, const Point& b) {
  134.     if (a.x==b.x&&a.y==b.y){
  135.         return true;
  136.     }
  137.     else{
  138.         return false;
  139.     }
  140. }
Success #stdin #stdout 0s 5360KB
comments (?)
stdin
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3 1 4 2
2 1 4 1
3 1 4 2
2 0 4 2
3 1 4 2
3 2 4 3
3 1 4 2
3 2 3 2
stdout
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(0,0)
兩線重疊
沒有交點
兩點重疊
https://ideone.com/FqghOQ
language:
C++ (gcc 8.3)
created:
1 year ago
visibility:
public

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