Hdu 1086 (judging segment intersection)

Portal: You can Solve a Geometry problem too

Test instructions: For n line segments, determine the number of points to intersect.

Analysis: To determine the intersection template problem, rapid rejection of the experimental principle is that each segment represents the vector and one end of the line segment and the other segment of the two end of the two vectors of the cross product, if the line intersection then the other line segment two endpoints must be on both sides of the segment, Then this segment represents the vector must be clockwise to turn the counter-clockwise, the cross product must be less than or equal to 0, the same to another line segment such a judgment can be.

#include <algorithm>#include<cstdio>#include<cstring>#include<cmath>#defineN 1010#definePI ACOs (-1.0)using namespacestd;Const Doubleeps=1e-8;intSgnDoublex) {    if(Fabs (x) <eps)return 0; if(x<0)return-1; return 1;}structpoint{Doublex, y; Point () {}, point (Double_x,Double_y): X (_x), Y (_y) {} pointoperator-(ConstPoint &a)Const    {        returnPoint (x-a.x,y-a.y); }    Double operator*(ConstPoint &a)Const    {        returna.x*x+a.y*y; }    Double operator^(ConstPoint &a)Const    {        returnx*a.y-y*a.x; }};structline{Point S,e; Line () {} line (point _s,point _e): S (_s), E (_e) {}};BOOLsegcrossseg (line l1,line L2) {returnMax (l1.s.x,l1.e.x)>=min (l2.s.x,l2.e.x) &&Max (l2.s.x,l2.e.x)>=min (l1.s.x,l1.e.x) &&Max (L1.S.Y,L1.E.Y)>=min (L2.S.Y,L2.E.Y) &&Max (L2.S.Y,L2.E.Y)>=min (L1.S.Y,L1.E.Y) &&SGN (L1.s-L1.E) ^ (L2.S-L1.E)) *SGN ((l1.s-l1.e) ^ (L2.E-L1.E)) <=0&&SGN (L2.s-L2.E) ^ (L1.S-L2.E)) *SGN ((l2.s-l2.e) ^ (L1.E-L2.E)) <=0;} Line l[ the];intMain () {intN;  while(SCANF ("%d",&N), N) { for(intI=1; i<=n;i++)        {            Doublea,b,c,d; scanf ("%LF%LF%LF%LF",&a,&b,&c,&d); L[i]=Line (point (b), point (C,d)); }        intans=0;  for(intI=1; i<=n;i++)             for(intj=i+1; j<=n;j++)            if(Segcrossseg (L[i],l[j])) ans++; printf ("%d\n", ans); }}

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Hdu 1086 (judging segment intersection)