Segment Intersection POJ 1066

1 //segment intersection POJ 10662 //idea: Direct enumeration of each endpoint and end point into a line segment, judging the number of intersections with the remaining segments3 4 //#include <bits/stdc++.h>5#include <iostream>6#include <cstdio>7#include <cstdlib>8#include <algorithm>9#include <vector>Ten#include <math.h> One using namespacestd; A #defineLL Long Long -typedef pair<int,int>PII; - Const intINF =0x3f3f3f3f; the ConstLL MOD =100000000LL; - Const intN = the; - #defineCLC (A, B) memset (A,b,sizeof (a)) - Const DoubleEPS = 1e-8; + voidFre () {freopen ("In.txt","R", stdin);} - voidFreout () {freopen ("OUT.txt","W", stdout);} +InlineintRead () {intx=0, f=1;CharCh=getchar (); while(ch>'9'|| ch<'0') {if(ch=='-') f=-1; Ch=getchar ();} while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; Ch=getchar ();}returnx*F;} A  at intSgnDoublex) { -     if(Fabs (x) < EPS)return 0; -     if(X <0)return-1; -     Else return 1; - } - structpoint{ in     Doublex, y; - Point () {} toPoint (Double_x,Double_y) { +x = _x;y =_y; -     } thePointoperator-(ConstPoint &b)Const{ *         returnPoint (X-b.x,y-b.y); $     }Panax Notoginseng     Double operator^(ConstPoint &b)Const{ -         returnx*b.y-y*b.x; the     } +     Double operator*(ConstPoint &b)Const{ A         returnx*b.x + y*b.y; the     } + }; -  $ structline{ $ Point s,e; -     intInx; - Line () {} the Line (Point _s,point _e) { -S=_s;e=_e;Wuyi     } the }; -  Wu Line Line[n]; - BOOLInter (line L1,line L2) { About     return  $Max (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) && ASGN ((l2.s-l1.s) ^ (L1.E-L1.S)) *SGN ((l2.e-l1.s) ^ (L1.E-L1.S)) <=0&& +SGN ((L1.S-L2.S) ^ (L2.E-L2.S)) *SGN ((L1.E-L2.S) ^ (L2.E-L2.S)) <=0; the } -  $Vector<line>list; the BOOLCMP (line l1,line L2) { the     returnl1.inx<L2.inx; the } the  -Point p[ the]; in intMain () { the     intN; the      while(~SCANF ("%d",&N)) { About      for(intI=1; i<=n;i++){ the         DoubleX1,y1,x2,y2; thescanf"%LF%LF%LF%LF",&x1,&y1,&x2,&y2); theline[i]=Line (Point (X1,y1), point (X2,y2)); +p[i*2-1]=Point (x1,y1); -p[i*2]=Point (x2,y2); the     }Bayi     Doublex1,y1; thescanf"%LF%LF",&x1,&y1); thePoint s=Point (x1,y1); -     intans=inf; -      for(intI=1; i<=2*n;i++){ the        inttem=0; the         for(intj=1; j<=n;j++){ the           if(Inter (line (S,p[i]), Line[j])) thetem++; -        } theans=min (ans,tem); the     } the     94 Point p1; theP1=point (0,0); the     inttem=0; the      for(intI=1; i<=n;i++){98         if(Inter (line (S,P1), Line[i])) Abouttem++; -     }101Ans=min (ans,tem+1);102     103P1=point (0, -);104tem=0; the      for(intI=1; i<=n;i++){106         if(Inter (line (S,P1), Line[i]))107tem++;108     }109Ans=min (ans,tem+1); the 111P1=point ( -,0); thetem=0;113      for(intI=1; i<=n;i++){ the         if(Inter (line (S,P1), Line[i])) thetem++; the     }117Ans=min (ans,tem+1);118 119P1=point ( -, -); -tem=0;121      for(intI=1; i<=n;i++){122         if(Inter (line (S,P1), Line[i]))123tem++;124     } theAns=min (ans,tem+1);126printf"Number of doors =");127printf"%d\n", ans); -     }129     return 0; the}

Segment Intersection POJ 1066