1 // 线段相交 poj 1066
2 // 思路：直接枚举每个端点和终点连成线段，判断和剩下的线段相交个数
3
4 // #include <bits/stdc++.h>
5 #include <iostream>
6 #include <cstdio>
7 #include <cstdlib>
8 #include <algorithm>
9 #include <vector>
10 #include <math.h>
11 using namespace std;
12 #define LL long long
13 typedef pair<int,int> pii;
14 const int inf = 0x3f3f3f3f;
15 const LL MOD =100000000LL;
16 const int N =110;
17 #define clc(a,b) memset(a,b,sizeof(a))
18 const double eps = 1e-8;
19 void fre() {freopen("in.txt","r",stdin);}
20 void freout() {freopen("out.txt","w",stdout);}
21 inline int read() {int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch=getchar();}while(ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();}return x*f;}
22
23 int sgn(double x){
24 if(fabs(x) < eps)return 0;
25 if(x < 0)return -1;
26 else return 1;
27 }
28 struct Point{
29 double x,y;
30 Point(){}
31 Point(double _x,double _y){
32 x = _x;y = _y;
33 }
34 Point operator -(const Point &b)const{
35 return Point(x - b.x,y - b.y);
36 }
37 double operator ^(const Point &b)const{
38 return x*b.y - y*b.x;
39 }
40 double operator *(const Point &b)const{
41 return x*b.x + y*b.y;
42 }
43 };
44
45 struct Line{
46 Point s,e;
47 int inx;
48 Line(){}
49 Line(Point _s,Point _e){
50 s=_s;e=_e;
51 }
52 };
53
54 Line line[N];
55 bool inter(Line l1,Line l2){
56 return
57 max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&
58 max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&
59 max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&
60 max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&
61 sgn((l2.s-l1.s)^(l1.e-l1.s))*sgn((l2.e-l1.s)^(l1.e-l1.s)) <= 0 &&
62 sgn((l1.s-l2.s)^(l2.e-l2.s))*sgn((l1.e-l2.s)^(l2.e-l2.s)) <= 0;
63 }
64
65 vector<Line> list;
66 bool cmp(Line l1,Line l2){
67 return l1.inx<l2.inx;
68 }
69
70 Point p[110];
71 int main(){
72 int n;
73 while(~scanf("%d",&n)){
74 for(int i=1;i<=n;i++){
75 double x1,y1,x2,y2;
76 scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
77 line[i]=Line(Point(x1,y1),Point(x2,y2));
78 p[i*2-1]=Point(x1,y1);
79 p[i*2]=Point(x2,y2);
80 }
81 double x1,y1;
82 scanf("%lf%lf",&x1,&y1);
83 Point s=Point(x1,y1);
84 int ans=inf;
85 for(int i=1;i<=2*n;i++){
86 int tem=0;
87 for(int j=1;j<=n;j++){
88 if(inter(Line(s,p[i]),line[j]))
89 tem++;
90 }
91 ans=min(ans,tem);
92 }
93
94 Point p1;
95 p1=Point(0,0);
96 int tem=0;
97 for(int i=1;i<=n;i++){
98 if(inter(Line(s,p1),line[i]))
99 tem++;
100 }
101 ans=min(ans,tem+1);
102
103 p1=Point(0,100);
104 tem=0;
105 for(int i=1;i<=n;i++){
106 if(inter(Line(s,p1),line[i]))
107 tem++;
108 }
109 ans=min(ans,tem+1);
110
111 p1=Point(100,0);
112 tem=0;
113 for(int i=1;i<=n;i++){
114 if(inter(Line(s,p1),line[i]))
115 tem++;
116 }
117 ans=min(ans,tem+1);
118
119 p1=Point(100,100);
120 tem=0;
121 for(int i=1;i<=n;i++){
122 if(inter(Line(s,p1),line[i]))
123 tem++;
124 }
125 ans=min(ans,tem+1);
126 printf("Number of doors = ");
127 printf("%d\n",ans);
128 }
129 return 0;
130 }
