POJ 2653 && HDU 1147 Pick-up sticks（计算几何）

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Description
有n根木条，一根一根的往一个坐标系上丢(给出木条两点的坐标)，问最后不被覆盖的木条有哪些，即丢的木条如果和前面丢的木条交叉的话，就会覆盖前面那根木条
Input
多组用例，每组用例第一行为木条数n，之后n行每行四个浮点数表示木条两端坐标，以n=0结束输入
Output
对于每组用例，输出最后不被覆盖的木条
Sample Input
5
1 1 4 2
2 3 3 1
1 -2.0 8 4
1 4 8 2
3 3 6 -2.0
3
0 0 1 1
1 0 2 1
2 0 3 1
0
Sample Output
Top sticks: 2, 4, 5.
Top sticks: 1, 2, 3.
Solution
枚举每根木条，如果它后面的木条有与其相交的则将其标记，最后没被标记的就是未被覆盖的木条
Code

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const double eps = 1e-8;#define maxn 111111int 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;    Line(){}    Line(Point _s,Point _e)    {        s=_s;e=_e;    }};//判断线段相交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;}Line line[maxn];bool flag[maxn];int main(){    int n;    double x1,x2,y1,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 res=true;        for(int i=1;i<=n;i++)            if(flag[i])            {                if(res)                    res=false;                else                    printf(", ");                printf("%d",i);            }        printf(".\n");    }    return 0;}

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