思路：枚举四边墙的门的中点，与终点连成一条线段，判断与其相交的线段的个数。最小的加一即为答案。
我是傻逼，一个数组越界调了两个小时。
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
const int N=;
const double eps=1e-;
int sgn(double x){
if(fabs(x)<eps) return ;
if(x>) return ;
return -;
}
struct point{
double x,y;
point(){}
point(double x_,double y_){
x=x_,y=y_;
}
point operator -(const point &b)const{
return point(x-b.x,y-b.y);
}
double operator *(const point &b)const{
return x*b.x+y*b.y;
}
double operator ^(const point &b)const{
return x*b.y-y*b.x;
}
}po[N];
struct line{
point s,e;
line(){}
line(point s_,point e_){
s=s_,e=e_;
}
}li[N];
double cal(point p0,point p1,point p2){//叉积
return (p1-p0)^(p2-p0);
}
int xj(line a,line b){//判断两线段是否相交
point A=a.s,B=a.e,C=b.s,D=b.e;
return
max(A.x,B.x)>=min(C.x,D.x) &&
max(C.x,D.x)>=min(A.x,B.x) &&
max(A.y,B.y)>=min(C.y,D.y) &&
max(C.y,D.y)>=min(A.y,B.y) &&
sgn(cal(C,A,B))*sgn(cal(D,A,B))<= &&
sgn(cal(A,C,D))*sgn(cal(B,C,D))<=;
}
int p[][];
int main(){
int n,i,j,js;
double x1,y1,x2,y2;
while(~scanf("%d",&n)){
memset(p,,sizeof(p));js=;
for(i=;i<=n;i++){
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
p[(int)x1][(int)y1]=p[(int)x2][(int)y2]=;
li[i]=line(point(x1,y1),point(x2,y2));
}
scanf("%lf%lf",&x1,&y1);
point P=point(x1,y1);
p[][]=;
p[][]=;
p[][]=;
p[][]=;
int x0,y0;
for(y0=,i=;i<=;i++){
if(p[][i]){
po[++js]=point(,(i+y0)/2.0);
y0=i;
}
}
for(x0=,i=;i<=;i++){
if(p[i][]){
po[++js]=point((i+x0)/2.0,);
x0=i;
}
}
for(y0=,i=;i<=;i++){
if(p[][i]){
po[++js]=point(,(i+y0)/2.0);
y0=i;
}
}
for(x0=,i=;i<=;i++){
if(p[i][]){
po[++js]=point((i+x0)/2.0,);
x0=i;
}
}
int ans=;
for(i=;i<=js;i++){
line L=line(P,po[i]);
int cnt=;
for(j=;j<=n;j++){
if(xj(L,li[j])){
cnt++;
}
}
if(cnt<ans) ans=cnt;
}
printf("Number of doors = %d\n",ans+);
}
return ;
}