一個體育生的程式設計之路（一）補充

2010年7月做的筆試題：（下邊是我自己當時做出來的答案，當時隻會JAVA，如果有更好的方法，請不吝賜教！）

第一個題目

在一個坐标系中已知兩個矩形的左下頂點和右上頂點的坐标，如果兩個矩形有重疊區域，求出重疊區域矩形的左下頂點坐标和右上頂點坐标。

public class Mytest{        public static void main(String[] args){               float x1=1.0f,y1=1.0f;      //第一個矩形左下頂點               float x2=3.0f,y2=3.0f;      //第一個矩形右上頂點               float x3=2.0f,y3=2.0f;      //第二個矩形左下頂點               float x4=4.0f,y4=4.0f;      //第二個矩形右上頂點               float x5,y5;   //重合區域矩形左下頂點               float x6,y6;   //重合區域矩形右上頂點               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("兩個矩形有重疊區域。\n 重疊區域的矩形坐标為：");                      System.out.println("x5= " + x5 + ", y5=" + y5);                      System.out.println("x6= " + x6 + ", y6=" + y6);               }        } }

第二個題目

第二個題目是已經三角形ABC三個頂點的坐标，A(x1,y1),   B(x2,y2),  C(x3,y3) ,   和另外一點P(x,y) 的坐标，怎麼判斷在三角形内還是外

雖然做出來了，但是不高效，現在我也不知道高效的方法是什麼？哪位大俠告訴小弟。

方法一：求面積之和

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);               //判斷               if(sabc-(spab+spbc+spac)<0.0000001){                      System.out.println("P點在這個三角形内部或者邊上");               }else {                      System.out.println("P點不在這個三角形外部");               }        } }

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) ;               //判斷               if(apb+apc+bpc-Math.PI*2<0.0000001){                      System.out.println("P點在這個三角形内部或者邊上");               }else {                      System.out.println("P點在這個三角形外部");               }        } }