HDU You can Solve a Geometry Problem too【线段交点判断】

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

分析：判断两直线是否相交：

分两步：

①：快速排斥实验

   设以线段P1P2为对角线的矩形为R，以线段Q1Q2为对角线的矩形为T，如果R和T不相交，显然两线段不会相交。

②：跨立实验

   如果两直线相交，则两直线必然相互跨立对方。若P1P2跨立Q1Q2，则矢量（P1-Q1）和（P2-Q1）位于矢量(Q2-Q1)的两侧，

 即(P1-Q1)X(Q2-Q1)*(P2-Q1)X(Q2-Q1)<0. 上式可改写为

     (P1-Q1)X(Q2-Q1)*(Q2-Q1)X(P2-Q1)>0. 当(P1-Q1)X(Q2-Q1)=0 时，说明(P1-Q1)和(Q2-Q1)共线，但因为已经通过快速排斥实验，所以

 P1一定在线段Q1Q2上，同理，(Q2-Q1)X(P2-Q1)=0 说明P2一定在直线Q1Q2上。所以判断P1P2跨立Q1Q2的依据是：

(P1 - Q1) X (Q2 - Q1) * (Q2 - Q1) X (P2 -Q1)>=0

 

code :

#include<stdio.h>
struct node
{
    double x0,y0;
    double x1,y1;
}q[101];
int p(struct node a,struct node b)
{
    double k1=(a.x0-b.x0)*(b.y1-b.y0)-(a.y0-b.y0)*(b.x1-b.x0);
    double k2=(b.x1-b.x0)*(a.y1-b.y0)-(b.y1-b.y0)*(a.x1-b.x0);
    if(k1*k2>=0)
    return 1;
    return 0;
}
int main()
{
    int t,tot,i,j;
    while(scanf("%d",&t),t)
    {
        tot=0;
        for(i=0;i<t;i++)
            scanf("%lf%lf%lf%lf",&q[i].x0,&q[i].y0,&q[i].x1,&q[i].y1);
        for(i=0;i<t-1;i++)
            for(j=i+1;j<t;j++)
                tot+=(p(q[i],q[j])&&p(q[j],q[i]));
        printf("%d\n",tot);
    }
    return 0;
}