poj 2653 线段相交裸题（解题报告）

#include<stdio.h>
#include<math.h>

const double eps=1e-8;
int n;
int cmp(double x)
{

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

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

Line line [(int)1e5+5];
double xmult(Point p0,Point p1,Point  p2)
{
    return (p1-p0)^(p2-p0);
}

bool seg_seg(Line l1,Line l2)
{
    return cmp(xmult(l1.s,l2.s,l2.e))*cmp(xmult(l1.e,l2.s,l2.e))<=0&&cmp(xmult(l2.e,l1.s,l1.e))*cmp(xmult(l2.s,l1.s,l1.e))<=0;
}

bool check(int num)
{
    for(int j=num+1;j<n;j++)
    {
        if(seg_seg(line[num],line[j])==true)
            return false;
    }
    return true;
}

int a[1005];
int main()
{
    double x1,x2,y1,y2;
    while(scanf("%d",&n),n)
    {
        int t=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            line[i]=Line(Point(x1,y1),Point(x2,y2));
        }
        for(int i=0;i<n;i++)
            if(check(i))
            a[t++]=i;
        printf("Top sticks: ");
        for(int i=0;i<t;i++)
        {
            printf("%d",a[i]+1);
            if(i<t-1)
                printf(", ");

        }
        printf(".
");
    }
}

思路：TOP STICK ：压在最上边的木棍，从最先放的木棍开始遍历，如果在其后边的木棍不压在它身上，即其后的木棍不与其相交（线段相交），则这个木棍就是top stick