10445 - Make Polygon

[Ghost77 dimen]

I use formula to get the max and min angle.

But it doesn't work.

How to calculate for higher precision, thanks?

My formula are

1. cosine law

2. vector law

Or the test data has tricks is the main reason of wrong answer.

Please let me know, thanks.

[RuiFerreira]

I'm having the same problem

[cytse]

i think the polygon maybe concave

[turuthok]

I almost gave up during the contest for this ... but I noticed that there are some important points that led to my AC here:

-> The points can be in clockwise or counter-clockwise order.
-> There might be > 180 angles.
-> I don't care whether or not we have 3 consecutive points being colinear, I treated the angle as 180 degree angle instead of ignoring it.
-> The polygon is not necessarily convex.

-turuthok-

[benetin]

My program handles all of these, but still...

I found sometimes the error appears exactly on the 7th digit. I got 179.999999 instead of 180.000000.

Got WA even if I artifically fix the 180 degree situation. So I wonder the other answers may also wrong due to the precision.?

Can anybody help me?

[Pupirev Sergei]

I can't find a mistake!
Help, please.

Code: Select all

#include <iostream.h>
#include <math.h>
#include <iomanip.h>
long p(long x1,long y1,long x2,long y2)
	{
	if (x1*y2-x2*y1>0) return 1;
	return 0;
	}
double l(long x1,long y1)
	{
	return sqrt(double(x1*x1)+double(y1*y1));
	}
double angle(long x1,long y1,long x2,long y2)
	{
	double a=double(x1*x2+y1*y2)/double(l(x1,y1)*l(x2,y2));
	a=acos(a);
	return (90.0*a)/acos(0.0);
	}
int main()
	{
	long n,i;
	long x[20],y[20];
	cin>>n;
	while (n>=3)
		{
		double mina=-1,maxa=-1;
		for (i=0;i<n;i++)
			cin>>x[i]>>y[i];
		int min=0;
		for (i=0;i<n;i++)
			if (x[min]>x[i]) min=i;
		long k;
		if (y[(min+1)%n]>y[(min+n-1)%n]) k=-1;
		else k=1;
		for (i=0;i<n;i++)
			{
			long x1=x[min%n],y1=y[min%n];
			long x2=x[(min+k+n)%n],y2=y[(min+k+n)%n];
			long x3=x[(min+2*k+n)%n],y3=y[(min+2*k+n)%n];
			if (p(x2-x1,y2-y1,x3-x2,y3-y2)==1)
				{
				double teca=angle(x1-x2,y1-y2,x3-x2,y3-y2);
				if (mina==-1 || teca<mina) mina=teca;
				if (maxa==-1 || teca>maxa) maxa=teca;
				}
			else
				{
				double teca=360.0-angle(x1-x2,y1-y2,x3-x2,y3-y2);
				if (mina==-1 || teca<mina) mina=teca;
				if (maxa==-1 || teca>maxa) maxa=teca;
				}
			min+=k;
			if (min<0) min+=n;
			}
		cout.setf(ios::fixed);
		cout.setf(ios::showpoint);

		cout<<setprecision(6)<<mina<<" "<<maxa<<"\n";
		cin>>n;
		}
	return 0;
	}

[turuthok]

Could you try this input (both polygons are the same):

Code: Select all

4
0 0
5 5
0 10
0 9
4
0 10
0 9
0 0
5 5
0

I hope it conforms to the input-description.

-turuthok-

[Pupirev Sergei]

Thanks!
I got AC!

[Dominik Michniewski]

It's possible, that input looks like:
4 0 0 1 0 1 1000 1 0
??
My program fails only on such case I think ....

DM

[marong]

I don't know why I got WA
can Anyone give me the input/output file?

Code: Select all

# include <stdio.h>
# include <math.h>
# define MaxN 200
# define PI 3.14159265359

struct Tpoint
{ long x; long y;}
Polygon[MaxN];

typedef struct Tpoint line;
typedef struct Tpoint point;

int N;
int Clockwise;

double LineDis(line a)
{
 return(sqrt(a.x*a.x+a.y*a.y));
}

long Cross_Product(line a,line b)
{
 return(a.x*b.y-b.x*a.y);
}

double arg(line a,line b)
{
 if (Cross_Product(a,b)*Clockwise<0)
    return(
	   PI-acos((a.x*b.x+a.y*b.y)/LineDis(a)/LineDis(b))
          );
 else
    return(
	   PI+acos((a.x*b.x+a.y*b.y)/LineDis(a)/LineDis(b))
          );
}

int Inline(point a,point b,point c)
{
 line t1,t2;

 t1.x=a.x-b.x;
 t1.y=a.y-b.y;
 t2.x=b.x-c.x;
 t2.y=b.y-c.y;
 return(Cross_Product(t1,t2)==0);
}

long Area(void)
{
  int i;
  long s=0;

  for (i=0;i<N;i++)
    s+=Cross_Product(Polygon[i],Polygon[(i+1)%N]);
  return(s/2);
}

void ReadInput(void)
{
 int i;
 for (i=0;i<N;i++)
    scanf("%ld %ld",&Polygon[i].x,&Polygon[i].y);
}

main()
{
 int i;
 double MaxArg,MinArg;
 double CurArg;
 line t1,t2;

  while (scanf("%d",&N),N>=3)
    {
      ReadInput();
      Clockwise=(Area()<0);
      if (Clockwise==0) Clockwise=-1;
      MaxArg=0;
      MinArg=360;
      for (i=0;i<N;i++)
	if (!Inline(Polygon[i],Polygon[(i+1)%N],Polygon[(i+N-1)%N]))
	{
          t1.x=Polygon[i].x-Polygon[(i+N-1)%N].x;
	  t1.y=Polygon[i].y-Polygon[(i+N-1)%N].y;
	  t2.x=Polygon[(i+1)%N].x-Polygon[i].x;
	  t2.y=Polygon[(i+1)%N].y-Polygon[i].y;
	  CurArg=arg(t1,t2)/PI*180;
          if (MaxArg<CurArg)
             MaxArg=CurArg;
	  if (MinArg>CurArg)
             MinArg=CurArg;
        }
      printf("%0.6f %0.6f\n",MinArg,MaxArg);
    }
}

[rakeb]

5
-1 0
1 0
1 10
0 5
-1 5

the output is:

11.309932 258.690068

i think it will help u . try to assign pi=acos(-1) or 2*acos(0). in this problem points may be given in clockwise or anticlockwise order. so u have to handle this case.

[lowai]

will the polygon be concave?

[marong]

Sorry,but I got the exactly correct answer about the case,

Have you any other case?

[rakeb]

6
0 0
1 0
1 10
0 5
-1 10
-1 0

20
0 0
1 0
2 0
3 0
4 0
5 0
6 0
7 0
8 0
9 0
9 1
8 1
7 1
6 1
5 1
4 1
3 1
2 1
1 1
0 1

outputs are:

11.309932 337.380135
90.000000 180.000000

to lowai:

will the polygon be concave?

u read the problem statement carefully. then u will find it.

[marong]

I puzzled.
if there are three point p p[i+1] p[i+2] in one line,
Should I ignore it or just count it as 180?

I has submitted two programme of these,
neither is ac.