线段相交 poj 1066

// 线段相交 poj 1066
// 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

// #include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <math.h>
using namespace std;
#define LL long long
typedef pair<int,int> pii;
const int inf = 0x3f3f3f3f;
const LL MOD =100000000LL;
const int N =110;
#define clc(a,b) memset(a,b,sizeof(a))
const double eps = 1e-8;
void fre() {freopen("in.txt","r",stdin);}
void freout() {freopen("out.txt","w",stdout);}
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;}

int sgn(double x){
    if(fabs(x) < eps)return 0;
    if(x < 0)return -1;
    else return 1;
}
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.y - y*b.x;
    }
    double operator *(const Point &b)const{
        return x*b.x + y*b.y;
    }
};

struct Line{
    Point s,e;
    int inx;
    Line(){}
    Line(Point _s,Point _e){
        s=_s;e=_e;
    }
};

Line line[N];
bool inter(Line l1,Line l2){
    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) &&
        sgn((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;
}

vector<Line> list;
bool cmp(Line l1,Line l2){
    return l1.inx<l2.inx;
}

Point p[110];
int main(){
    int n;
    while(~scanf("%d",&n)){
    for(int i=1;i<=n;i++){
        double x1,y1,x2,y2;
        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
        line[i]=Line(Point(x1,y1),Point(x2,y2));
        p[i*2-1]=Point(x1,y1);
        p[i*2]=Point(x2,y2);
    }
    double x1,y1;
    scanf("%lf%lf",&x1,&y1);
    Point s=Point(x1,y1);
    int ans=inf;
    for(int i=1;i<=2*n;i++){
       int tem=0;
       for(int j=1;j<=n;j++){
          if(inter(Line(s,p[i]),line[j]))
            tem++;
       }
       ans=min(ans,tem);
    }
    
    Point p1;
    p1=Point(0,0);
    int tem=0;
    for(int i=1;i<=n;i++){
        if(inter(Line(s,p1),line[i]))
          tem++;
    }
    ans=min(ans,tem+1);
    
    p1=Point(0,100);
    tem=0;
    for(int i=1;i<=n;i++){
        if(inter(Line(s,p1),line[i]))
          tem++;
    }
    ans=min(ans,tem+1);

    p1=Point(100,0);
     tem=0;
    for(int i=1;i<=n;i++){
        if(inter(Line(s,p1),line[i]))
          tem++;
    }
    ans=min(ans,tem+1);

    p1=Point(100,100);
    tem=0;
    for(int i=1;i<=n;i++){
        if(inter(Line(s,p1),line[i]))
          tem++;
    }
    ans=min(ans,tem+1);
    printf("Number of doors = ");
    printf("%d
",ans);
    }
    return 0;
}