LA 4064 (counting polar angles sort) Magnetic Train Tracks

This question is similar to UVa11529.

Enumerates a central point and then sorts it by a polar angle, counting the number of triangles that are obtuse at this point, and subtracting the answer with C (n, 3).

In addition encounter right triangle situation is very egg pain, can use an EPS, not too troublesome words to use integer vector do dot product.

1#include <cstdio>2#include <cmath>3#include <algorithm>4 using namespacestd;5 6typedefLong LongLL;7 Const intMAXN = -+Ten;8 Const DoublePI = ACOs (-1.0);9 Const DoubleEPS = 1e-9;Ten  One struct Point A { -     Doublex, y; - Point () {} thePoint (DoubleXDoubley): x (x), Y (y) {} - }P[MAXN], P2[MAXN]; -  -Pointoperator- (Constpoint& A,Constpoint&B) + { -     returnPoint (a.x-b.x, a.y-b.y); + } A  at DoubleANG[MAXN *2]; -  -LL-Inline C3 (intN) {return(LL) n * (n1) /2* (n2) /3; } -  - intMain () - { in     //freopen ("In.txt", "R", stdin); -     intN, Kase =0; to  +      while(SCANF ("%d", &n) = =1&&N) -     { the          for(inti =0; I < n; i++) scanf ("%LF%LF", &p[i].x, &p[i].y); *  $LL cnt =0;Panax Notoginseng          for(inti =0; I < n; i++) -         { the             intK =0; +              for(intj =0; J < N; J + +)if(J! =i) A             { theP2[K] = p[j]-P[i]; +ANG[K] =atan2 (P2[K].Y, p2[k].x); -Ang[k + N-1] = Ang[k] + PI *2.0; $k++; $             } -K =2*n-2; -Sort (ang, Ang +k); the             intL, R1 =0, R2 =0; -              for(L =0; L < n1; l++)Wuyi             { the                 DoubleB1 = ang[l] + PI/2; -                 DoubleB2 = Ang[l] +PI; Wu                  while(Ang[r1] <= b1-eps) r1++; -                  while(ANG[R2] < B2) r2++; AboutCNT + = R2-R1; $             } -         } -LL ans = C3 (n)-CNT; -printf"Scenario%d:\nthere is%lld sites for making valid tracks\n", ++Kase, ans); A     } +  the     return 0; -}

code June

LA 4064 (counting polar angles sort) Magnetic Train Tracks