第三次作业
本人代码水平十分有限，仅供参考，有错误请指出
java源码：
package robathomework3;
import java.lang.Math;
//点类
class point {
double x = 0;
double y = 0;
point(double x, double y) {
this.x = x;
this.y = y;
}
void print() {
System.out.println("(" + x + "," + y + ")");
}
}
//圆类，给定圆心点和半径R即可确定唯一一个圆
class Circle {
private double x;
private double y;
private double r;
//构造函数，用point对象和一个double变量初始化
public Circle(point point, double R) {
this.x = point.x;
this.y = point.y;
this.r = R;
}
public double getX() {
return x;
}
public void setX(double x) {
this.x = x;
}
public double getY() {
return y;
}
public void setY(double y) {
this.y = y;
}
public double getR() {
return r;
}
public void setR(double r) {
this.r = r;
}
public void print() {
System.out.println("(" + this.x + "," + this.y + ")" + this.r);
}
}
//计算类
class Function {
// 初始化，防止Nullpointer异常
private Circle c1 = new Circle(new point(0, 0), 0);
private Circle c2 = new Circle(new point(0, 0), 0);
// deg1/2是储存以自身为中心的极角
private double deg1;
private double deg2;
//构造函数，截取字符串，初始化各个变量
public Function(String s) {
String[] s1 = s.substring(1, s.indexOf(")")).split(" ");
String[] s2 = s.substring(s.indexOf(")") + 3, s.length() - 1).split(" ");
this.c1 = new Circle(Getpoint(s1[0]), Double.parseDouble(s1[1]));
this.c2 = new Circle(Getpoint(s2[0]), Double.parseDouble(s2[1]));
this.deg1 = Double.parseDouble(s1[2]);
this.deg1 = Double.parseDouble(s2[2]);
}
//通过字符标志返回点
public point Getpoint(String p) {
switch (p) {
case "P1":
return new point(-52.5, -32);
case "P2":
return new point(-52.5, 32);
case "P3":
return new point(52.5, 32);
case "P4":
return new point(52.5, -32);
case "P5":
return new point(0, -32);
case "P6":
return new point(0, 32);
case "P7":
return new point(-30, -7);
case "P8":
return new point(-30, 7);
case "P9":
return new point(30, 7);
case "P10":
return new point(30, -7);
case "C":
return new point(0, 0);
case "G1":
return new point(-52.5, 0);
case "G2":
return new point(52.5, 0);
case "T1":
return new point(2, 0);
case "T2":
return new point(0, 1);
default:
return new point(0, 0);
}
}
//判断两个圆有几个交点
public int Judeg() {
double d = Math.sqrt(
(c1.getX() - c2.getX()) * (c1.getX() - c2.getX()) + (c1.getY() - c2.getY()) * (c1.getY() - c2.getY()));
if (this.c1.getR() + this.c2.getR() < d) {
return 0;
} else if (this.c1.getR() + this.c2.getR() == d) {
return 1;
} else {
return 2;
}
}
//关键方法，如果有两个交点，判断哪个交点是机器人所在位置
//主要思路：假定机器人所在位置是坐标原点，建立一个直角坐标系，通过给定的两个点，可分别求出其在上述坐标轴中，与机器人所在点形成的连线与X轴的夹角
//此夹角配合给定的自身极角，就可以分别计算出头部在此坐标系中面向的角度。
//理论上，正确的位置，两个定点求出的头部方向应该相同，但是，计算难免有误差，因此我想到，用这两个 头部角度做差
//理论上的正确位置的差，就算计算有误差，其值应该也比非正确位置上的差小。这样，通过比较差值，便可选择到正确的点。
public double IsMyPoint(point point) {
double deg_1 = 0, deg_2 = 0;
deg_1 = Math.acos(Math.abs((c1.getX() - point.x) / Math
.sqrt((c1.getX() - point.x) * (c1.getX() - point.x) + (c1.getY() - point.y) * (c1.getY() - point.y))));
deg_1 = Math.toDegrees(deg_1);
deg_2 = Math.acos(Math.abs((c2.getX() - point.x) / Math
.sqrt((c2.getX() - point.x) * (c2.getX() - point.x) + (c2.getY() - point.y) * (c2.getY() - point.y))));
deg_2 = Math.toDegrees(deg_2);
//以机器人自身为原点建立坐标系，为了保证角度计算的一致性，我们可以讨论顶点落在的象限，就可以统一角度
//这里我规定了逆时针方向为增，同时所有象限的角都为正，即从x轴正方向开始，0-360.
//而机器人的视野角度是规定的逆时针为负，因此可以看到，后面是减去机器人的自身极角的
if (point.x <= c1.getX() && point.y <= c1.getY()) {
;
} else if (point.x > c1.getX() && point.y <= c1.getY()) {
deg_1 = 180 - deg_1;
} else if (point.x > c1.getX() && point.y > c1.getY()) {
deg_1 = 180 + deg_1;
} else {
deg_1 = 360 - deg_1;
}
if (point.x <= c2.getX() && point.y <= c2.getY()) {
;
} else if (point.x > c2.getX() && point.y <= c2.getY()) {
deg_2 = 180 - deg_2;
} else if (point.x > c2.getX() && point.y > c2.getY()) {
deg_2 = 180 + deg_2;
} else {
deg_2 = 360 - deg_2;
}
deg_1 = deg_1 - this.deg1;
deg_2 = deg_2 - this.deg2;
// System.out.println(deg_1);
// System.out.println(deg_2);
if (deg_1 < 0) {
deg_1 = 360 - deg_1;
}
if (deg_2 < 0) {
deg_2 = 360 - deg_2;
}
return Math.abs(deg_1 - deg_2);
}
//计算函数，计算交点，配合上述选择交点的函数，返回一个正确位置
public point GetIntersection() {
double a = 0, b = 0, c = 0;
// double d =Math.sqrt((c1.getX()-c2.getX())+(c1.getY()-c2.getY()));
double x1 = 0, y1 = 0, x2 = 0, y2 = 0;
if (c1.getY() != c2.getY()) {
double A = (c1.getX() * c1.getX() - c2.getX() * c2.getX() + c1.getY() * c1.getY() - c2.getY() * c2.getY()
+ c2.getR() * c2.getR() - c1.getR() * c1.getR()) / (2 * (c1.getY() - c2.getY()));
double B = (c1.getX() - c2.getX()) / (c1.getY() - c2.getY());
a = 1 + B * B;
b = -2 * (c1.getX() + (A - c1.getY()) * B);
c = c1.getX() * c1.getX() + (A - c1.getY()) * (A - c1.getY()) - c1.getR() * c1.getR();
if (Judeg() == 2) {
x1 = (-b + Math.sqrt(b * b - 4 * a * c)) / (2 * a);
x2 = (-b - Math.sqrt(b * b - 4 * a * c)) / (2 * a);
y1 = A - B * x1;
y2 = A - B * x2;
if (IsMyPoint(new point(x1, y1)) < IsMyPoint(new point(x2, y2))) {
return new point(x1, y1);
} else {
return new point(x2, y2);
}
} else if (Judeg() == 1) {
return new point(-b / (2 * a), A - B * x1);
} else {
System.out.println("两个圆不相交");
return null;
}
} else if (c1.getX() != c2.getX()) {
// 当y1=y2时，x的两个解相等
x1 = x2 = (c1.getX() * c1.getX() - c2.getX() * c2.getX() + c2.getR() * c2.getR() - c1.getR() * c1.getR())
/ (2 * (c1.getX() - c2.getX()));
a = 1;
b = -2 * c1.getY();
c = c1.getY() * c1.getY() - c1.getR() * c1.getR() + (x1 - c1.getX()) * (x1 - c1.getX());
if (Judeg() == 2) {
y1 = (-b + Math.sqrt(b * b - 4 * a * c)) / (2 * a);
y2 = (-b - Math.sqrt(b * b - 4 * a * c)) / (2 * a);
if (IsMyPoint(new point(x1, y1)) < IsMyPoint(new point(x2, y2))) {
return new point(x1, y1);
} else {
return new point(x2, y2);
}
} else if (Judeg() == 1) {
y1 = y2 = -b / (2 * a);
return new point(x1, y1);
} else {
System.out.println("两个圆不相交");
return null;
}
} else {
System.out.println("两圆心重合");
return null;
}
}
}
public class RobatHomework3 {
public static void main(String args[]) {
// 返回的是一个点对象
Function function = new Function("(C 2 0) (T1 2 60)");
function.GetIntersection().print();
Function function2 = new Function("(C 2 0) (T1 2 -60)");
function2.GetIntersection().print();
Function function3 = new Function("(P7 7 0) (P8 7 -180)");
function3.GetIntersection().print();
Function function4 = new Function("(P8 22 0) (P7 15 30)");
function4.GetIntersection().print();
Function function5 = new Function("(P6 30 0) (P3 22.5 180)");
function5.GetIntersection().print();
}
}
运行效果图：
输入数据分别为：
(C 2 0) (T1 2 60) 其中T1(2,0)
(C 2 0) (T1 2 -60)
(P7 7 0) (P8 7 -180)
(P8 22 0) (P7 15 30)
(P6 30 0) (P3 22.5 180)
所用公式：
)
