[poj2653]Pick-up sticks

题目大意：给定一系列线段，以及放在平面上的顺序，给出没有被其他覆盖的线段。

解题关键：线段相交的判断。

满足两个条件即可：快速排斥实验、跨立实验。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<iostream> 
using namespace std;
typedef long long ll;
const double eps=1e-8;
int sgn(double x){
    if(fabs(x)<eps)return 0;
    else 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;
    Line(){}
    Line(Point _s,Point _e){s=_s;e=_e;}
};
//判断线段相交，模板 
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;
} 

const int MAXN=100010;
Line line[MAXN];
bool flag[MAXN];
int main(){
    int n;
    double x1,y1,x2,y2;
    while(scanf("%d",&n),n){
        for(int i=1;i<=n;i++){
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            line[i]=Line(Point(x1,y1),Point(x2,y2));
            flag[i]=true;
        }
        for(int i=1;i<=n;i++){
            for(int j=i+1;j<=n;j++)
                if(inter(line[i],line[j])){
                    flag[i]=false;
                    break;
                }
        }
        printf("Top sticks: ");
        bool first=true;
        for(int i=1;i<=n;i++)
            if(flag[i]){//只是为了控制格式 
                if(first)first=false;
                else printf(", ");
                printf("%d",i);
            }
        printf(".\n");
    }
    
    return 0;
}