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 ",ans); 128 } 129 return 0; 130 }