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-;
 struct Point{
     double x,y;
     Point(){}
     Point(double x0,double y0):x(x0),y(y0){}
 }s[],e[];
 int ans[],st[],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 ;//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=;i<=n;i++) ans[i]=;
         for (int i=;i<=n;i++){
             scanf("%lf%lf%lf%lf",&s[i].x,&s[i].y,&e[i].x,&e[i].y);
         }
         for (int i=;i<=n;i++)
          for (int j=i+;j<=n;j++)
           if (judge(s[i],e[i],s[j],e[j]))
          { ans[i]=;
            break;
          }
         int top=;
         for (int i=;i<=n;i++)
          if (!ans[i])
           st[++top]=i;
         printf("Top sticks:");
         for (int i=;i<top;i++)
          printf(" %d,",st[i]);
         printf(" %d.\n",st[top]);
     }
 }