java 判断一个点是否在三角形内
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Finding whether a point is within a triangle
提问by Lish
I have been on this for hours, attempting different methods looking at just about every question. Perhaps I have it completely wrong, but I feel that I have my math of it correct, but no matter what numbers I input, I get the same output. My code is off somewhere and I have to turn it in by midnight.
我已经研究了几个小时，尝试不同的方法来解决几乎每个问题。也许我完全错了，但我觉得我的数学是正确的，但是无论我输入什么数字，我都会得到相同的输出。我的代码在某个地方关闭，我必须在午夜之前打开它。
It is the all so fun: Find if a point is within a triangle code. (for beginners)
最有趣的是：查找一个点是否在三角形代码内。（对新手而言）
import java.util.Scanner;
public class PointsTriangle {
// checks if point entered is within the triangle
//given points of triangle are (0,0) (0,100) (200,0)
public static void main (String [] args) {
//obtain point (x,y) from user
System.out.print("Enter a point's x- and y-coordinates: ");
Scanner input = new Scanner(System.in);
double x = input.nextDouble();
double y = input.nextDouble();
//find area of triangle with given points
double ABC = ((0*(100-0 )+0*(0 -0)+200*(0-100))/2.0);
double PAB = ((x*(0 -100)+0*(100-y)+0 *(y- 0))/2.0);
double PBC = ((x*(100-0 )+0*(0 -y)+200*(y-100))/2.0);
double PAC = ((x*(0 -100)+0*(100-y)+200*(y- 0))/2.0);
boolean isInTriangle = PAB + PBC + PAC == ABC;
if (isInTriangle)
System.out.println("The point is in the triangle");
else
System.out.println("The point is not in the triangle");
}//end main
}//end PointsTriangle
采纳答案by Floris
If you draw a picture, you can see the point has to satisfy simple inequalities (below / above / to the right of certain lines). Whether "on the edge" is in or out I will leave up to you:
如果您画图，您可以看到该点必须满足简单的不等式（某些线的下方/上方/右侧）。无论“在边缘”是进还是出，我都将留给您：
Y > 0 (above the X axis)
X > 0 (to the right of the Y axis)
X + 2* Y < 200 (below the hypotenuse)
Write an if statement around these three and you're done:
围绕这三个写一个 if 语句，你就完成了：
if( (y > 0) && (x > 0) && (x + 2*y < 200) )
System.out.println("The point is in the triangle");
else
System.out.println("The point is not in the triangle");
回答by Thai Tran
You placed the wrong value order into your formula; therefore, the result is wrong. If the 3 vertices are as the following
您在公式中放置了错误的值顺序；因此，结果是错误的。如果3个顶点如下
A(x1, y1) B(x2, y2), C(x3, y3)
then the area is calculated as
然后面积计算为
double ABC = Math.abs (x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2)) / 2;
After that, you just replace each vertex with the input point, we will have the following triangles: PBC, APC, ABP.
之后，您只需将每个顶点替换为输入点，我们将得到以下三角形：PBC、APC、ABP。
Put everything together, we will have the correct one
把所有的东西放在一起，我们会有正确的
int x1 = 0, y1 = 0;
int x2 = 0, y2 = 100;
int x3 = 200, y3 = 0;
// no need to divide by 2.0 here, since it is not necessary in the equation
double ABC = Math.abs (x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2));
double ABP = Math.abs (x1 * (y2 - y) + x2 * (y - y1) + x * (y1 - y2));
double APC = Math.abs (x1 * (y - y3) + x * (y3 - y1) + x3 * (y1 - y));
double PBC = Math.abs (x * (y2 - y3) + x2 * (y3 - y) + x3 * (y - y2));
boolean isInTriangle = ABP + APC + PBC == ABC;
