在c ++中绘制多边形并使用面积公式在输出中显示其面积。 [英] draw polygon in c++ and show its area in output using polygon area formula.

问题描述

就像我想使用graphics.h在c ++中绘制一个多边形。然后询问坐标和边。 x1，x2，y1，y2。然后绘制多边形并显示其面积和周长作为输出。使用他们的mathamatical公式。

我做到了这一点。

like i want to draw a polygon in c++ using graphics.h . then ask for coordinates and sides. x1, x2 ,y1 ,y2. then draw the polygon and show its area and perimeter as output. using their mathamatical formulas.
I have done this.

#include <iostream.h>
 #include <graphics.h>
 #include    <conio.h>
 #include     <math.h>

 void show_screen( );

 void Polygon(const int,const int []);

 void Line(const int,const int,const int,const int);

 int main( )
    {
       int driver=VGA;
       int mode=VGAHI;

       int n=0;

       do
      {
	 show_screen( );

	 gotoxy(12,13);
	 cout<<"Enter the number of points : ";
	 cin>>n;

	 int *coordinates=new int[(n*2)];

	 for(int count=0;count<n;count++)>
	{
	   gotoxy(8,18);
	   cout<<"Coordinates of Point-"<<(count+1)<<" (x"<<(count+1)<<",y"<<(count+1)<<") :";


	   gotoxy(12,21);
	   cout<<"Enter the value of x"<<(count+1)<<" = ";
	   cin>>coordinates[(count*2)];

	   gotoxy(12,22);
	   cout<<"Enter the value of y"<<(count+1)<<" = ";
	   cin>>coordinates[((count*2)+1)];

	   gotoxy(8,18);
	   cout<<"                                            ";

	   gotoxy(12,21);
	   cout<<"                                            ";

	   gotoxy(12,22);
	   cout<<"                                            ";
	 }

	 initgraph(&driver,&mode,"c:\\tc\\Bgi");

	 setcolor(15);
	   Polygon(n,coordinates);

	     delete coordinates;

	 setcolor(15);
	   outtextxy(110,460,"Press <enter> to continue or any other key to exit.");

	 int key=int(getch( ));

	 if(key!=13)
	break;
      }
       while(1);

       return 0;
    }

 void Polygon(const int n,const int coordinates[])
    {
       if(n>=2)
      {
	 Line(coordinates[0],coordinates[1],
			 coordinates[2],coordinates[3]);

	 for(int count=1;count<(n-1);count++)
	Line(coordinates[(count*2)],coordinates[((count*2)+1)],
			coordinates[((count+1)*2)],
			coordinates[(((count+1)*2)+1)]);
      }
    }

void Line(const int x_1,const int y_1,const int x_2,const int y_2)
    {
       int color=getcolor( );

       int x1=x_1;
       int y1=y_1;

       int x2=x_2;
       int y2=y_2;

       if(x_1>x_2)
      {
	 x1=x_2;
	 y1=y_2;

	 x2=x_1;
	 y2=y_1;
      }

       int dx=abs(x2-x1);
       int dy=abs(y2-y1);
       int inc_dec=((y2>=y1)?1:-1);

       if(dx>dy)
      {
	 int two_dy=(2*dy);
	 int two_dy_dx=(2*(dy-dx));
	 int p=((2*dy)-dx);

	 int x=x1;
	 int y=y1;

	 putpixel(x,y,color);

	 while(x<x2)>
	{
	   x++;

	   if(p<0)
	      p+=two_dy;

	   else
	      {
	     y+=inc_dec;
	     p+=two_dy_dx;
	      }

	   putpixel(x,y,color);
	}
      }

       else
      {
	 int two_dx=(2*dx);
	 int two_dx_dy=(2*(dx-dy));
	 int p=((2*dx)-dy);

	 int x=x1;
	 int y=y1;

	 putpixel(x,y,color);

	 while(y!=y2)
	{
	   y+=inc_dec;

	   if(p<0)
	      p+=two_dx;

	   else
	      {
	     x++;
	     p+=two_dx_dy;
	      }

	   putpixel(x,y,color);
	}
      }
    }

void show_screen( )
    {
       restorecrtmode( );
       textmode(C4350);

       textbackground(1);
       cprintf(" Polygon ");
       textbackground(8);


       gotoxy(1,2);
    }

推荐答案

假设您可以从用户那里收集输入，那么绘制多边形是微不足道的：

call

Assuming you are able collecting input from the user, then drawing the polygon is trivial:
call

moveto(x0,y0); lineto(x1,y1);...;lineto(x0,y0);

计算周长相对容易，因为例如

Computing the perimeter is relatively easy, since, for instance

l01 = sqrt((x1-x0)*(x1-x0)+(y1-y0)*(y1-y0));

给出了连接点 0 的长度到点 1 。

计算该区域有点棘手，您可以在这里找到一种有趣的方法：不规则多边形区域 [ ^ ]。

gives the length of the side connecting point 0 to point 1.
Computing the area is a bit trickier, here you may find an interesting approach: "Area of Irregular Polygons"[^].

我们不做你的作业：它是有原因的。它就是为了让你思考你被告知的事情，并试着理解它。它也在那里，以便您的导师可以识别您身体虚弱的区域，并将更多的注意力集中在补救措施上。



亲自尝试，你可能会发现它不是和你想的一样困难！



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