HDU1086：You can Solve a Geometry Problem too(线段相交模板)

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

 

 

这道题要用到线段相交的知识，虽然线段相交咋看听容易，但真正要把代码写出来还是不简单的

反正我在百度搜索的时候要用到两个知识点，建议有兴趣的同学可以去认真的理解下

 

#include 
#include 
#include 
#include 
using namespace std;

struct point
{
    double x,y;
};
point a[105][2];

double fan(double x,double y)
{
    return x>y?x:y;
}

double fin(double c,double d)
{
    return c=fin(a.x,b.x)&&m2>=fin(a.y,b.y)&&m2<=fan(a.y,b.y)&&m1<=fan(c.x,d.x)&&m1>=fin(c.x,d.x)&&m2>=fin(c.y,d.y)&&m2<=fan(c.y,d.y))
            return 1;
    }
    if(c.x==d.x&&a.x!=b.x)
    {
        double m1=c.x;
        double m2=a.y+(b.y-a.y)*(c.x-a.x)/(b.x-a.x);
        if(m1<=fan(a.x,b.x)&&m1>=fin(a.x,b.x)&&m2>=fin(a.y,b.y)&&m2<=fan(a.y,b.y)&&m1<=fan(c.x,d.x)&&m1>=fin(c.x,d.x)&&m2>=fin(c.y,d.y)&&m2<=fan(c.y,d.y))
            return 1;
    }
    double k1=(b.y-a.y)/(b.x-a.x);
    double k2=(d.y-c.y)/(d.x-c.x);
    double m1,m2,x,y;
    if(k1==k2)  return 0;
    else
    {
        m1=a.y-k1*a.x;
        m2=c.y-k2*c.x;
        x=(m1-m2)/(k2-k1);
        y=k1*x+m1;
        if(x<=fan(a.x,b.x)&&x>=fin(a.x,b.x)&&y>=fin(a.y,b.y)&&y<=fan(a.y,b.y)&&x<=fan(c.x,d.x)&&x>=fin(c.x,d.x)&&y>=fin(c.y,d.y)&&y<=fan(c.y,d.y))
            return 1;
    }
    return 0;
}

int main()
{
    int cas = 1;
    int n,i,j;
    while(~scanf("%d",&n),n)
    {
        int cnt = 0;
        for(i = 0;i