一个体育生的编程之路（一）补充

2010年7月做的笔试题：（下边是我自己当时做出来的答案，当时只会JAVA，如果有更好的方法，请不吝赐教！）

第一个题目

在一个坐标系中已知两个矩形的左下顶点和右上顶点的坐标，如果两个矩形有重叠区域，求出重叠区域矩形的左下顶点坐标和右上顶点坐标。

public class Mytest{        public static void main(String[] args){                      float x1=1.0f,y1=1.0f;      //第一个矩形左下顶点               float x2=3.0f,y2=3.0f;      //第一个矩形右上顶点               float x3=2.0f,y3=2.0f;      //第二个矩形左下顶点               float x4=4.0f,y4=4.0f;      //第二个矩形右上顶点                 float x5,y5;   //重合区域矩形左下顶点               float x6,y6;   //重合区域矩形右上顶点                 x5 = x1<=x3?x3:x1;               y5 = y1<=y3?y3:y1;               x6 = x2<=x4?x2:x4;               y6 = y2<=y4?y2:y4;               if(x5<x6&&y5<y6){                      System.out.println("两个矩形有重叠区域。\n 重叠区域的矩形坐标为：");                      System.out.println("x5= " + x5 + ", y5=" + y5);                      System.out.println("x6= " + x6 + ", y6=" + y6);               }        } }

 

 

第二个题目

第二个题目是已经三角形ABC三个顶点的坐标，A(x1,y1),   B(x2,y2),  C(x3,y3) ,   和另外一点P(x,y) 的坐标，怎么判断在三角形内还是外

 

虽然做出来了，但是不高效，现在我也不知道高效的方法是什么？哪位大侠告诉小弟。

方法一：求面积之和

public class Mytest02 {        public static void main(String[] args){               double x1=1.0, y1=1.0 ;           //A               double x2=3.0, y2=1.0 ;           //B               double x3=2.0, y3=3.0 ;   //C                        double x=2.0, y=2.0 ;  //P                 double pa = Math.sqrt((x-x1)*(x-x1)+(y-y1)*(y-y1));               double pb = Math.sqrt((x-x2)*(x-x2)+(y-y2)*(y-y2));               double pc = Math.sqrt((x-x3)*(x-x3)+(y-y3)*(y-y3));                      double ab = Math.sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));               double ac = Math.sqrt((x1-x3)*(x1-x3)+(y1-y3)*(y1-y3));               double bc = Math.sqrt((x2-x3)*(x2-x3)+(y2-y3)*(y2-y3));                      double abc = (ab+bc+ac)/2 ;                      double pab = (pa+pb+ab)/2;               double pbc = (pb+pc+bc)/2;               double pac = (pa+pc+ac)/2; System.out.println(abc); System.out.println(pab); System.out.println(pbc); System.out.println(pac);               double sabc = Math.sqrt(abc*(abc-ab)*(abc-bc)*(abc-ac));                      double spab = Math.sqrt(pab*(pab-pa)*(pab-pb)*(pab-ab));               double spbc = Math.sqrt(pbc*(pbc-pb)*(pbc-pc)*(pbc-bc));               double spac = Math.sqrt(pac*(pac-pa)*(pac-pc)*(pac-ac)); System.out.println(sabc); System.out.println(spab); System.out.println(spbc); System.out.println(spac);               //判断               if(sabc-(spab+spbc+spac)<0.0000001){                      System.out.println("P点在这个三角形内部或者边上");               }else {                      System.out.println("P点不在这个三角形外部");               }        }       }

 

 

方法二：求角度之和

public class Mytest03 {        public static void main(String[] args){               double x1=1.0, y1=1.0 ;           //A               double x2=3.0, y2=1.0 ;           //B               double x3=2.0, y3=3.0 ;   //C                        double x=2.0, y=2.0 ;  //P                 double pa = Math.sqrt((x-x1)*(x-x1)+(y-y1)*(y-y1));               double pb = Math.sqrt((x-x2)*(x-x2)+(y-y2)*(y-y2));               double pc = Math.sqrt((x-x3)*(x-x3)+(y-y3)*(y-y3));                      double ab = Math.sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));               double ac = Math.sqrt((x1-x3)*(x1-x3)+(y1-y3)*(y1-y3));               double bc = Math.sqrt((x2-x3)*(x2-x3)+(y2-y3)*(y2-y3));        double apb = Math.acos((pa*pa+pb*pb-ab*ab)/(2*pa*pb)) ; double apc = Math.acos((pa*pa+pc*pc-ac*ac)/(2*pa*pc)) ; double bpc = Math.acos((pb*pb+pc*pc-bc*bc)/(2*pb*pc)) ; System.out.println(apb) ; System.out.println(apc) ; System.out.println(bpc) ; System.out.println(Math.PI*2) ;               //判断               if(apb+apc+bpc-Math.PI*2<0.0000001){                      System.out.println("P点在这个三角形内部或者边上");               }else {                      System.out.println("P点在这个三角形外部");               }        }   }