453  Intersecting Circles
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 Experienced poster
 Posts: 105
 Joined: Wed May 25, 2005 7:23 am
I get the same output which adrian posted but still WA.
Who can help me?Point out my fault.thanks!
The following is my code:
Who can help me?Point out my fault.thanks!
The following is my code:
Code: Select all
ACed at last
Last edited by pineapple on Mon Apr 30, 2007 5:21 am, edited 1 time in total.
Its really tough to get this one accepted. And finding bug is impossible. Try changing these...
And eps should be positive and small. 1e7 or 1e8 will do. Try changing these.
Code: Select all
if(arbr==0.0)
Dont use this, try to use
if( fabs(arbr) < eps )
Ami ekhono shopno dekhi...
HomePage
HomePage
453 PE
Many threads to this problem are about Wrong Answer. My first submit was WA too, but next 4 submissions I had PE. I don't understand, why
I think, that messages ("NO INTERSECTION" or "THE CIRCLES ARE THE SAME") and coordinates of the points (with parenthesis) I print correct, without any blanks. I think about extra lines in the end of input. I avoid to write "0.000", I write "0.000". But still PE. Blank lines shouldn't be here...
Have someone idea, where may be the reason of PE? Many solution have PE in this problem. {198 AC/ 145 PE in this time}
Thanks in advance!
Please sorry for my bad english.
I think, that messages ("NO INTERSECTION" or "THE CIRCLES ARE THE SAME") and coordinates of the points (with parenthesis) I print correct, without any blanks. I think about extra lines in the end of input. I avoid to write "0.000", I write "0.000". But still PE. Blank lines shouldn't be here...
Have someone idea, where may be the reason of PE? Many solution have PE in this problem. {198 AC/ 145 PE in this time}
Thanks in advance!
Please sorry for my bad english.
I got PE several times, too. Then I changed the 'eps' value and got accepted.
Ami ekhono shopno dekhi...
HomePage
HomePage
I've tried with Epsilon = 1e4 and 1e8 and got PE in both cases.
What might be wrong?
The PE ratio is also quite high.
This problem has a special corrector (probably because of the accuracies?)
What might be wrong?
The PE ratio is also quite high.
This problem has a special corrector (probably because of the accuracies?)
Jose Santos
http://ctp.di.fct.unl.pt/~jcas
http://ctp.di.fct.unl.pt/~jcas
453 WA
Everybody is talking about getting PE, but man I cannot remove WA talk about PE later. Anyways looked for Adrian's inputs, but they are not any more available neither from Stefen site nor Adrians. Somebody please come to rescue, I'm going crazy already tried arnd 20 submissions with different combinations. Heeeeeeeeeelllllllllpppppppppp.
Code: Select all
#include <stdio.h>
#include <math.h>
#define EPS 1e4
#define EQUAL(a,b) (fabs((a)(b))<EPS)
#define ROUND(a) ((long)(((a)+5e4)*1000))
#define SQR(a) ((a)*(a))
#define ZERO(a) (fabs(a)<EPS)
void main()
{
double x1, y1, r1, x2, y2, r2, D, A, ix1, iy1, ix2, iy2;
//freopen("453.in", "r+", stdin);
//printf("%lf", EPS);
while ( scanf("%lf %lf %lf %lf %lf %lf", &x1, &y1, &r1, &x2, &y2, &r2) == 6 )
{
D = SQR(x1  x2) + SQR(y1  y2);
D = sqrt(D);
A = (D + r1 + r2) * (D + r1  r2) * (D  r1 + r2) * (D + r1 + r2);
if ( EQUAL(x1, x2) && EQUAL(y1, y2) && EQUAL(r1, r2) )
{
puts("THE CIRCLES ARE THE SAME");
}
else if (D < fabs(r1  r2)  D > (r1 + r2) )
//else if ((r1 + r2) < D )
{
puts("NO INTERSECTION");
}
else
{
A = (sqrt(A) / 4.0);
ix1 = ix2 = ((x1 + x2) / 2)  (((x1  x2) * (SQR(r1)SQR(r2))) / (2 * SQR(D)));
ix1 = ix1 + ((2 * (y1  y2) * A) / SQR(D));
ix2 = ix2  ((2 * (y1  y2) * A) / SQR(D));
iy1 = iy2 = ((y1 + y2) / 2)  (((y1  y2) * (SQR(r1)SQR(r2))) / (2 * SQR(D)));
iy1 = iy1  ((2 * (x1  x2) * A) / SQR(D));
iy2 = iy2 + ((2 * (x1  x2) * A) / SQR(D));
ix1 = (ZERO(ix1) ? 0 : ix1);
iy1 = (ZERO(iy1) ? 0 : iy1);
ix2 = (ZERO(ix2) ? 0 : ix2);
iy2 = (ZERO(iy2) ? 0 : iy2);
//if ( ROUND(ix1) == ROUND(ix2) && ROUND(iy1) == ROUND(iy2) )
if ( EQUAL(ix1, ix2) && EQUAL(iy1, iy2) )
{
printf("(%.3lf,%.3lf)\n", ix1, iy1);
}
else
{
if ( (ix1 < ix2)  (EQUAL(ix1, ix2) && (iy1 < iy2)) )
printf("(%.3lf,%.3lf)(%.3lf,%.3lf)\n", ix1, iy1, ix2, iy2);
else
printf("(%.3lf,%.3lf)(%.3lf,%.3lf)\n", ix2, iy2, ix1, iy1);
}
}
}
}
Really tough problem...
My code passed the test case provided by the other post...yet, WA
Roundoff problem is difficult to tackle... for this online judge.
I think dealing with the roundoff error "correctly" requires some "luck".
BTW, why I can't search the posts of this problem after I entered "453"?
My code passed the test case provided by the other post...yet, WA
Roundoff problem is difficult to tackle... for this online judge.
I think dealing with the roundoff error "correctly" requires some "luck".
BTW, why I can't search the posts of this problem after I entered "453"?
Code: Select all
#include <iostream>
#include <cstdio>
#include <math.h>
#define EPSILON 1e4
using namespace std;
double xc1, xc2, yc1, yc2;
double r1, r2;
inline bool EQUAL(double X, double Y){
return fabs(XY)<EPSILON;
}
inline double zz(double X){
if(fabs(X)<EPSILON) return 0.000;
else return X;
}
void printPair(double X, double Y);
int main(){
while(scanf("%lf", &xc1)!=EOF){
scanf("%lf %lf", &yc1, &r1);
scanf("%lf %lf %lf", &xc2, &yc2, &r2);
if(EQUAL(xc1,xc2) && EQUAL(yc1,yc2) && EQUAL(r1,r2)){
if(EQUAL(r1,0.000)&&EQUAL(r2,0.000))
printf("(%.3lf,%.3lf)\n",xc1,yc1);
else
printf("THE CIRCLES ARE THE SAME\n");
continue;
}
double dx = xc1xc2;
double dy = yc1yc2;
double sqD = dx*dx+dy*dy;
double sumR = r1+r2;
double sqSumR = sumR*sumR;
double diffR = r1r2;
double sqDiffR = diffR*diffR;
if(EQUAL(sqD,sqSumR)){
// Touch externally
// Exactly 1 intersecting pt
double ratio = r2/sumR;
printPair(xc2+dx*ratio, yc2+dy*ratio);
printf("\n");
}else{
// Find Equation of Chord...
double d1=2*xc1, e1=2*yc1, f1=xc1*xc1+yc1*yc1r1*r1;
double d2=2*xc2, e2=2*yc2, f2=xc2*xc2+yc2*yc2r2*r2;
double A=d1d2, B=e1e2, C=f1f2;
double xFinal, yFinal;
// Touches internally
// Exactly 1 intersectin pt
if(EQUAL(sqD,sqDiffR)){
if(EQUAL(B,0.000)){
xFinal = C/A;
yFinal = yc1;
}else{
double AoverB = A/B, CoverB = C/B;
double a=1+AoverB*AoverB;
double b=2*xc1+2*AoverB*(yc1+CoverB);
xFinal = b/2/a;
yFinal = AoverB*xFinal  CoverB;
}
printPair(xFinal, yFinal);
printf("\n");
}else if(sqD > sqSumR  sqD < sqDiffR){
// No intersection
printf("NO INTERSECTION\n");
}else{
// Distinct intersections
if(EQUAL(B,0.000)){
printf("EQUAL B, DISTINCT\n");
xFinal = C/A;
double temp=xFinalxc1;
double c=r1*r1temp*temp;
double rDelta=sqrt(c);
printPair(xFinal, yc1rDelta);
printPair(xFinal, yc1+rDelta);
// printf("(%.3lf,%.3lf)\n", zz(xFinal), yc1+rDelta);
}else{
double AoverB = A/B, CoverB = C/B;
double a=1+AoverB*AoverB;
double b=2*xc1+2*AoverB*(yc1+CoverB);
double temp=CoverB+yc1; temp*=temp;
double c=xc1*xc1+tempr1*r1;
double DELTA = b*b4*a*c;
double rDelta = sqrt(DELTA);
xFinal = (brDelta)/2/a;
yFinal = AoverB*xFinal  CoverB;
printPair(xFinal, yFinal);
xFinal = (b+rDelta)/2/a;
yFinal = AoverB*xFinal  CoverB;
printPair(xFinal, yFinal);
}
printf("\n");
}
}
}
return 0;
}
void printPair(double X, double Y){
printf("(%.3lf,%.3lf)", zz(X), zz(Y));
}

 Experienced poster
 Posts: 110
 Joined: Tue May 06, 2008 2:18 pm
 Location: CSEDU, Bangladesh
 Contact:
Re: 453 argghhh
those who are still in trouble with this one: just make sure your formula is correct (no matter what you see in your pc)
I used eps = 1e4
carefully compare:
a > b should be a > b + eps
a < b should be a + eps < b
a == b should be fabs(a  b) < eps
for comparing if there is only one point, calculate two points as usual, then compare if they are same, don't use any other properties for checking if they are same, and don't forget to maintain the order. remember also that, there are circles of 0.00 radii.
before submitting, make eps = 1e4
I got PE with 1e5 and AC with 1e4, although, in my pc, 1e4 produces many incorrect answer where 1e5 produces more correct answers (according to IO provided with in this thread). got ac in 3 submissions...
nice geometry with frustrating IO set.
I used eps = 1e4
carefully compare:
a > b should be a > b + eps
a < b should be a + eps < b
a == b should be fabs(a  b) < eps
for comparing if there is only one point, calculate two points as usual, then compare if they are same, don't use any other properties for checking if they are same, and don't forget to maintain the order. remember also that, there are circles of 0.00 radii.
before submitting, make eps = 1e4
I got PE with 1e5 and AC with 1e4, although, in my pc, 1e4 produces many incorrect answer where 1e5 produces more correct answers (according to IO provided with in this thread). got ac in 3 submissions...
nice geometry with frustrating IO set.
You should not always say what you know, but you should always know what you say.

 New poster
 Posts: 15
 Joined: Thu Sep 02, 2010 3:10 pm
 Location: Dhaka,Bangladesh
 Contact:
Re: 453
gootsa,please check this testcase..
please ,send your new code to me,if you dont mind....
usuttra@yahoo.com
Code: Select all
input
1.0 1.0 5.0
1.0 1.0 10.0
output must be
(1.000,1.000)
usuttra@yahoo.com

 New poster
 Posts: 15
 Joined: Thu Sep 02, 2010 3:10 pm
 Location: Dhaka,Bangladesh
 Contact:
Re: 453 argghhh
I got AC on this problem.
check this test case given below.U will get AC also...
check this test case given below.U will get AC also...
Code: Select all
1.0 1.0 5.0
1.0 1.0 5.0
1.0 1.0 0.0
1.0 1.0 0.0
5.2 2.5 0.0
6.1 8.2 0.0
Re: 453 PE
Code: Select all
if(EQUAL(xc1,xc2) && EQUAL(yc1,yc2) && EQUAL(r1,r2))
453: Query
Why the output for the following input is "NO INTERSECTION"?
Can anybody prove mathematically?
I see distance of two centers is 0.280176, some of two radius is 1.44844. So there should be intersections.
Code: Select all
0.931547 0.869930 0.866543
0.674520 0.758415 0.581896
I see distance of two centers is 0.280176, some of two radius is 1.44844. So there should be intersections.