POJ 2653 Pick-up sticks(线段相交)

题意：给定n个木棍依次放下，要求最终判断没被覆盖的木棍是哪些。

思路:快速排斥以及跨立实验可以判断线段相交。

#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
const double eps=1e-10;
struct Point{
    double x,y;
    Point(){}
    Point(double x0,double y0):x(x0),y(y0){}
}s[200005],e[200005];
int ans[200005],st[200005],n;
double operator *(Point p1,Point p2){
    return p1.x*p2.y-p1.y*p2.x;
}
Point operator -(Point p1,Point p2){
    return Point(p1.x-p2.x,p1.y-p2.y);
}
bool judge(Point p1,Point p2,Point p3,Point p4){
    if (std::min(p1.x,p2.x)>std::max(p3.x,p4.x)||
        std::min(p1.y,p2.y)>std::max(p3.y,p4.y)||
        std::min(p3.x,p4.x)>std::max(p1.x,p2.x)||
        std::min(p3.y,p4.y)>std::max(p1.y,p2.y))
    return 0;//快速排斥实验
    double a,b,c,d;
    a=(p2-p1)*(p3-p1);
    b=(p2-p1)*(p4-p1);
    c=(p4-p3)*(p1-p3);
    d=(p4-p3)*(p2-p3);
    return (a*b<=eps)&&(c*d<=eps);//跨立实验
}
int main(){
    while (scanf("%d",&n)!=EOF&&n){
        for (int i=1;i<=n;i++) ans[i]=0;
        for (int i=1;i<=n;i++){
            scanf("%lf%lf%lf%lf",&s[i].x,&s[i].y,&e[i].x,&e[i].y);
        }
        for (int i=1;i<=n;i++)
         for (int j=i+1;j<=n;j++)
          if (judge(s[i],e[i],s[j],e[j]))
         { ans[i]=1;
           break;
         }
        int top=0; 
        for (int i=1;i<=n;i++)
         if (!ans[i])
          st[++top]=i;
        printf("Top sticks:");  
        for (int i=1;i<top;i++)
         printf(" %d,",st[i]);
        printf(" %d.
",st[top]);    
    }
}