Test instructions
There are n a circle of different sizes placed in a sequential order, and the length of all visible arcs is calculated.
Analysis:
This question should be an example of the Great White Book LA 2572 (the number of visible discs) Kanazawa, the overall framework is similar.
For each circle, the amplitude angle corresponding to the intersection of the other circles (converted to [0, 2π]) is ordered, and the end point of each arc is enumerated, and is visible if it is not covered by the circle that is placed behind it.
Attention:
Originally thought WA is the precision problem, later in the major small has been WA, here Precision EPS from 1e-11 to 1e-13 is no problem.
But an equal sign is added when judging whether the end of the arc is covered by a circle. That is, the 64th line of code is <= rather than <
The UVA data is really strong, Orz.
1#include <cstdio>2#include <cmath>3#include <vector>4#include <algorithm>5 using namespacestd;6 7 Const intMAXN = -+Ten;8 Const DoubleEPS = 1e- One;9 Const DoublePI = ACOs (-1.0);Ten Const DoubleTWO_PI =2.0*PI; One DoubleRADIUS[MAXN]; A - DoubleNormalizeangle (Doubleang) -{returnAng-two_pi*floor (ang/two_pi); } the - intDCMP (Doublex) - { - if(Fabs (x) < EPS)return 0; + returnX <0? -1:1; - } + A struct Point at { - Doublex, y; -Point (Doublex=0,Doubley=0): X (x), Y (y) {} - }P[MAXN]; - typedef point Vector; - in BOOL operator== (Constpoint& A,Constpoint&B) -{returnDCMP (a.x-b.x) = =0&& dcmp (a.y-b.y) = =0; } to +Pointoperator- (Constpoint& A,Constpoint&B) -{returnPoint (a.x-b.x, A.Y-b.y); } the * DoubleDot (Constpoint& A,Constpoint&B) ${returna.x*b.x + a.y*b.y;}Panax Notoginseng - DoubleLength (Constvector&A) the{returnsqrt (Dot (A, a));} + A DoubleAngle (Constvector&A) the{returnatan2 (A.Y, a.x);} + - voidGetccintersection (Constpoint& C1,DoubleR1,Constpoint& C2,DoubleR2, vector<Double>&rad) $ { $ DoubleD = Length (C1-C2); - if(DCMP (d) = =0)return; - if(DCMP (D-R1-R2) >0)return; the if(DCMP (D-fabs (R1-R2)) <0)return; - Wuyi Double Base= Angle (C2-C1); the DoubleAng = ACOs ((r1*r1 + d*d-r2*r2)/(2.0*r1*d)); -Rad.push_back (Normalizeangle (Base+ang)); WuRad.push_back (Normalizeangle (Base-ang)); - } About $ intN; - - BOOLIsVisible (Constpoint& C,intID) - { A for(inti = ID +1; I < n; ++i) + { the DoubleD = Length (C-p[i]); - if(DCMP (d-radius[i]) <=0)return false;//the key to this problem $ } the return true; the } the the intMainvoid) - { in //freopen ("10969in.txt", "R", stdin); the intT; thescanf"%d", &T); About while(t--) the { thescanf"%d", &n); the for(inti =0; I < n; ++i) scanf ("%LF%LF%LF", &radius[i], &p[i].x, &p[i].y); + - Doublesum =0.0; the for(inti =0; I < n; ++i)Bayi { thevector<Double>Rad; theRad.push_back (0.0); - Rad.push_back (TWO_PI); - for(intj =0; J < N; ++j) the { the if(j = = i)Continue; the getccintersection (P[i], radius[i], p[j], radius[j], RAD); the } - sort (Rad.begin (), Rad.end ()); the the for(intj =0; J < Rad.size ()-1; ++j) the {94 DoubleMID = (Rad[j] + rad[j +1]) /2; the DoubleAng = Rad[j +1] -Rad[j]; thePoint C (p[i].x + Radius[i]*cos (mid), P[i].y + radius[i]*sin (mid)); the if(IsVisible (C, i)) sum + = radius[i] *ang;98 } About } - 101printf"%.3f\n", sum);102 }103 104 return 0; the}
code June
UVa 10969 (cover problem between circle and round) Sweet Dream