Calculate geometry to determine intersection of segments
Note the sentence in the title: You could assume that there is no more than top sticks.
I think it is not described clearly, if not every time the throw is guaranteed not more than 1000, the problem is very difficult to do.
If each throw is guaranteed to be less than 1000 root at the top, then violence can be.
Open a vector to save each throw at the top of the stick, the next time you throw, as long as the removal of the vector and throw in the intersection of the stick, and then put the stick into the, and finally left is the answer
#include <cstdio>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<list>#include<algorithm>using namespacestd;Const intmaxn=100000+Ten;Const Doubleeps=1e-8;intTotp,totl;structpoint{Doublex; Doubley;} p[2*MAXN];structline{Point A; Point B; intID;} Line[maxn];vector<Line>V;vector<Line>:: Iterator Iter;vector<int>ans;intFLAG[MAXN];DoubleXmult (Point p1,point p2,point p0) {return(p1.x-p0.x) * (P2.Y-P0.Y)-(p2.x-p0.x) * (p1.y-p0.y);}intopposite_side (Point p1,point p2,point l1,point L2) {returnXmult (L1,P1,L2) *xmult (L1,P2,L2) <-EPS;}intintersect_ex (Point u1,point u2,point v1,point v2) {returnOpposite_side (U1,U2,V1,V2) &&opposite_side (V1,V2,U1,U2);}intMain () { while(~SCANF ("%d",&Totl)) { if(!TOTL) Break; TotP=0; for(intI=0; i<totl; i++) { Doublea,b,c,d; scanf ("%LF%LF%LF%LF",&a,&b,&c,&d); P[TOTP+0].x=A; P[TOTP+0].y=b; P[TOTP+1].x=C; P[TOTP+1].y=D; LINE[I].A=p[totp+0]; LINE[I].B=p[totp+1]; Line[i].id=i; TotP=totp+2; } v.clear (); for(intI=0; i<totl; i++) { for(Iter=v.begin (); iter!=v.end ();) { line now=*Iter; if(INTERSECT_EX (line[i].a,line[i].b,now.a,now.b)) Iter=v.erase (Iter); Elseiter++; } v.push_back (Line[i]); } ans.clear (); for(Iter=v.begin (); Iter!=v.end (); iter++) { line now=*Iter; Ans.push_back (now.id); } printf ("Top Sticks:"); for(intI=0; I<ans.size (); i++) {printf ("%d", ans[i]+1); if(I<ans.size ()-1) printf (", "); Elseprintf". \ n"); } } return 0;}
POJ 2653 Pick-up Sticks