HDU1086 You can Solve a Geometry Problem too

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10594    Accepted Submission(s): 5246

Problem Description

Many geometry（几何）problems were designed in the ACM/ICPC. And now, I also prepare a geometry problem for this final exam. According to the experience of many ACMers, geometry problems are always much trouble, but this problem is very easy, after all we are now attending an exam, not a contest :)
Give you N (1<=N<=100) segments（线段）, please output the number of all intersections（交点）. You should count repeatedly if M (M>2) segments intersect at the same point.

Note:
You can assume that two segments would not intersect at more than one point. 

 

Input

Input contains multiple test cases. Each test case contains a integer N (1=N<=100) in a line first, and then N lines follow. Each line describes one segment with four float values x1, y1, x2, y2 which are coordinates of the segment’s ending. 
A test case starting with 0 terminates the input and this test case is not to be processed.

 

Output

For each case, print the number of intersections, and one line one case.

 

Sample Input

2 0.00 0.00 1.00 1.00 0.00 1.00 1.00 0.00 3 0.00 0.00 1.00 1.00 0.00 1.00 1.00 0.000 0.00 0.00 1.00 0.00 0

 

Sample Output

1 3

     ACM暑假训练时做的题目，一个模板题判断两条线段是否相交，涉及到数学里的向量叉乘，判断条件如下：
     1.两条线段相交，一条线段的左右端点在分别另一条线段的左右两侧，向量积表示。
     2.特殊情况，一条线段的端点在另一条线段上，向量积是0，要进行特判。
    就是代码丑了点O__O "…

#include 
#include 
using namespace std;
struct node {
  double x1,y1,x2,y2;
  }c[102];
int t;
int cross(node a,node b);
int on(double a,double b,double c,double d,double e,double f);
int main()
{
    while(~scanf("%d",&t)&&t){
        for(int i = 0;i < t;i++)
            scanf("%lf%lf%lf%lf",&c[i].x1,&c[i].y1,&c[i].x2,&c[i].y2);
        long long int sum = 0;
        for(int i = 0;i < t;i++){
            for(int j = i+1;j < t;j++){
                if(cross(c[i],c[j]))
                    sum++;
            }
        }
        printf("%I64d\n",sum);
    }
    return 0;
}
int cross(node a,node b)

{
  node r,t,s,x,y,z;
  //以线段a的(x1,y1)作为起点
  r.x1 = b.x1-a.x1;
  r.y1 = b.y1-a.y1;
  t.x1 = a.x2-a.x1;
  t.y1 = a.y2-a.y1;
  s.x1 = b.x2-a.x1;
  s.y1 = b.y2-a.y1;
  //以线段b的(x1,y1)作为起点
  x.x1 = a.x1-b.x1;
  x.y1 = a.y1-b.y1;
  y.x1 = b.x2-b.x1;
  y.y1 = b.y2-b.y1;
  z.x1 = a.x2-b.x1;
  z.y1 = a.y2-b.y1;

  //向量积运算
  if(((r.x1*t.y1-t.x1*r.y1<0&&s.x1*t.y1-t.x1*s.y1>0)||(r.x1*t.y1-t.x1*r.y1>0&&s.x1*t.y1-t.x1*s.y1<0))
     &&((x.x1*y.y1-y.x1*x.y1<0&&z.x1*y.y1-y.x1*z.y1>0)||(x.x1*y.y1-y.x1*x.y1>0&&z.x1*y.y1-y.x1*z.y1<0)))
     return 1;

  //四个特判
  else if(r.x1*t.y1-t.x1*r.y1==0&&on(a.x1,a.y1,a.x2,a.y2,b.x1,b.y1))
    return 1;
  else if(s.x1*t.y1-t.x1*s.y1==0&&on(a.x1,a.y1,a.x2,a.y2,b.x2,b.y2))
    return 1;
  else if(x.x1*y.y1-y.x1*x.y1==0&&on(b.x1,b.y1,b.x2,b.y2,a.x1,a.y1))
    return 1;
  else if(z.x1*y.y1-y.x1*z.y1==0&&on(b.x1,b.y1,b.x2,b.y2,a.x2,a.y2))
    return 1;
  return 0;

}
int on(double a,double b,double c,double d,double e,double f)

{
    if(min(a,c)<=e&&e<=max(a,c)&&min(b,d)<=f&&f<=max(b,d))
       return 1;
  return 0;

}