在二維平面上，有兩個正方形，請找出一條直線，可以將這兩個正方形對半分

/**
 * 功能：在二維平面上，有兩個正方形，請找出一條直線，可以將這兩個正方形對半分。
 * 假定正方形的上下兩條邊與x軸平行。

 */

 

[java] view plain copy

 

1. /** 
2.  * 考慮： 
3.  * 線的準確含義，可能性有： 
4.  *      1）由斜率和y軸截距肯定； 
5.  *      2）由這條邊上的任意兩點肯定； 
6.  *      3）線段，以正方形的邊做爲起點和終點。 
7.  *  
8.  * 假設：這條線的端點應該落在正方形的邊上。 
9.  * 思路：要將兩個正方形對半分，這條線必須鏈接兩個正方形的中心點。 
10.  */  
11.   
12. public class Square {  
13.   
14.     //正方形的四條邊  
15.     int left;  
16.     int right;  
17.     int top;  
18.     int bottem;  
19.     int size;  
20.       
21.     public static void main(String[] args) {  
22.         // TODO Auto-generated method stub  
23.   
24.     }  
25.   
26.     //獲得正方形的中心點的位置  
27.     public Point getMiddle(){  
28.         return new Point((this.left+this.right)/2.0,(this.top+this.bottem)/2);  
29.     }  
30.       
31.     //返回線段mid1和mid2的線段與square2的邊相交的點，即從mid1到mid2畫一條線，一直延伸置碰到square2的靠外的那條邊。  
32.     public Point extend(Point mid1,Point mid2,int size){  
33.         //肯定線段mid1->mid2的方向  
34.         int xdir=mid1.x<mid2.x?1:-1;  
35.         int ydir=mid1.y<mid2.y?1:-1;  
36.           
37.         //若是mid1和mid2的x座標相同，計算斜率時，會拋出零異常，作特別處理  
38.         if(mid1.x==mid2.y)  
39.             return new Point(mid2.x,mid2.y+ydir*size/2.0);  
40.           
41.         //計算線段的斜率  
42.         double slope=(mid2.y-mid1.y)/(mid2.x-mid1.x);  
43.         double x1=0;  
44.         double y1=0;  
45.           
46.         //計算相交點，注意斜率的取值  
47.         if(Math.abs(slope)==1){  
48.             x1=mid2.x+xdir*size/2.0;  
49.             y1=mid2.y+ydir*size/2.0;  
50.         }else if(Math.abs(slope)<1){  
51.             x1=mid2.x+xdir*size/2.0;  
52.             y1=mid2.y+slope*ydir*(size/2.0);//注意方向  
53.         }else if(Math.abs(slope)>1){  
54.             x1=mid2.x+slope*xdir*(size/2.0);  
55.             y1=mid2.y+ydir*size/2.0;  
56.         }  
57.         return new Point(x1,y1);  
58.     }  
59.       
60.     public MyLine cut(Square other){  
61.         //計算兩個中心點之間的線段與正方形的邊相交的位置  
62.         Point point1=extend(this.getMiddle(),other.getMiddle(),other.size);  
63.         Point point2=extend(this.getMiddle(),other.getMiddle(),-other.size);  
64.         Point point3=extend(other.getMiddle(),this.getMiddle(),this.size);  
65.         Point point4=extend(other.getMiddle(),this.getMiddle(),-this.size);  
66.           
67.         //找出線段的起點和終點  
68.         Point start=point1;  
69.         Point end=point1;  
70.         Point[] points={point1,point2,point3};  
71.         for(int i=0;i<points.length;i++){  
72.             if(points[i].x<start.x||points[i].x==start.x&&points[i].y<start.y)  
73.                 start=points[i];  
74.             else if(points[i].x>end.x||points[i].x==end.x&&points[i].y>end.y)  
75.                 end=points[i];  
76.         }  
77.           
78.         return new MyLine(start,end);  
79.     }  
80.       
81. }  
82.   
83. class MyLine{  
84.     Point start;  
85.     Point end;  
86.     public MyLine(Point start,Point end){  
87.         this.start=start;  
88.         this.end=end;  
89.     }  
90. }  
91.   
92. class Point{  
93.     double x;  
94.     double y;  
95.     public Point(double x,double y){  
96.         this.x=x;  
97.         this.y=y;  
98.     }  
99. }

或者：

public static double[] getBipartition(Point[] A, Point[] B) {                                                    

    // 此題因爲正方形的上下兩條邊與x軸平行, 因此就不用直線相交去求中心點了                           

    // 求出第一個點不一樣的座標(由於點的順序多是混亂的)                                     

    double p1x1 = A[0].x;                                            

    double p1x2 = A[1].x == A[0].x ? A[2].x:A[1].x;                  

    double p1y1 = A[0].y;                                            

    double p1y2 = A[1].y == A[0].y ? A[2].y:A[1].y;                  

                                                                       

    // 求出第二個點不一樣的座標(由於點的順序多是混亂的)                                     

    double p2x1 = B[0].x;                                            

    double p2x2 = B[1].x == B[0].x ? B[2].x:B[1].x;                  

    double p2y1 = B[0].y;                                            

    double p2y2 = B[1].y == B[0].y ? B[2].y:B[1].y;                  

                                                                       

    // 平分線的兩個點                                                       

    double x1 = (double)(p1x2 + p1x1) / 2;                           

    double y1 = (double)(p1y2 + p1y1) / 2;                           

    double x2 = (double)(p2x1 + p2x2) / 2;                           

    double y2 = (double)(p2y1 + p2y2) / 2;                           

                                                                       

    double k = (double)(y2 - y1) / (x2 - x1);                        

    double b = y1 - k*x1;                                            

                                                                       

    return new double[] {k,b};                                       

}  

 

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

public class Bipartition {     public double[] getBipartition(Point[] A, Point[] B) {         // write code here         //平分直線一定通過正方形的中心點         int x1 = (A[0].x+A[2].x)/2;         int y1 = (A[0].y+A[2].y)/2;                   int x2 = (B[0].x+B[2].x)/2;         int y2 = (B[0].y+B[2].y)/2;                   double[] a = new double[2];         a[0]=(double)(y2-y1)/(double)(x2-x1);         a[1]=y2-a[0]*x2;         return a;     } }

 

public class Bipartition {

    public double[] getBipartition(Point[] A, Point[] B) {

        // write code here

        double Acenterx=0;//正方形A的中心x座標

        double Acentery=0;//正方形A的中心y座標

        

        for(int i=0;i<4;i++){

            Acenterx+=A[i].x;

            Acentery+=A[i].y;

        }

        Acenterx/=4.0;

        Acentery/=4.0;

        

        double Bcenterx=0;//正方形B的中心x座標

        double Bcentery=0;//正方形B的中心y座標

        for(int i=0;i<4;i++){

            Bcenterx+=B[i].x;

            Bcentery+=B[i].y;

        }

        Bcenterx/=4.0;

        Bcentery/=4.0;

        

        double[] result=new double[2];

        result[0]=(Bcentery-Acentery)/(Bcenterx-Acenterx);//求斜率

        result[1]=Acentery-result[0]*Acenterx;//求截距

        

        return result;

    }

}