poj2653 线段相交的判断

//poj2653
//题目意思：逐根放置不同长度的木棒，求哪几根是没有被盖住的;
//解题思路：判断线段相交（）。值得注意的是判断线段相交的顺序，先从前面的与后面的判断，
//若相交，对前面的木棒对应的状态设为false,并退出这根木棒的判断，否则会超时.

#include<iostream>
#include<cstdio>
#include<math.h>
#define eps 1e-8
struct Point{double x,y;};
bool zero(double x)
{
return x>0? x<eps: x>-eps;
}
double xmult(Point p1,Point p2,Point p0)
{
return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);
}
int dot_online_in(Point p,Point l1,Point l2)//点在线段上
{
return zero(xmult(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps;
}
int same_side(Point p1,Point p2,Point p3,Point p4)
{
return xmult(p1,p3,p4)*xmult(p2,p3,p4)>eps;
}
int intersect_in(Point p1,Point p2,Point p3,Point p4)//线段相交
{
if(!zero(xmult(p1,p2,p3))&&!zero(xmult(p1,p2,p4)))
return !same_side(p1,p2,p3,p4)&&!same_side(p3,p4,p1,p2);
return dot_online_in(p1,p3,p4)||dot_online_in(p2,p3,p4)
||dot_online_in(p3,p1,p2)||dot_online_in(p4,p1,p2);
}
bool f[100002];
Point p[100002][2];
int main()
{
int n;
while(scanf("%d",&n),n)
{
int i,j;
for(i=0;i<=n;i++)
f[i]=true;
for(i=1;i<=n;i++)
{
scanf("%lf%lf%lf%lf",&p[i][0].x,&p[i][0].y,&p[i][1].x,&p[i][1].y);
}
for(i=1;i<=n;i++)
{
for(j=i+1;j<=n;j++)
{
if(intersect_in(p[i][0],p[i][1],p[j][0],p[j][1])){f[i]=false;break;}
}
}
printf("Top sticks:");
for(j=1;j<=n;j++)
{
if(f[j])
{
printf(" %d",j);
if(j==n)printf(".\n");
else printf(",");
}
}
}
return 0;
}