HDU1086（判断两线段是否相交）

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

 

Author

lcy

判断两线段AB,CD相交的方法

1、判断A和B在CD两侧

2、判断C和D在AB两侧

主要利用ACxAB*ADxAB<=0

 CAxCD*CBxCD<=0

(其中x表示两个向量的叉乘，*表示普通的乘法)

具体的关于计算几何的问题可以参考
http://dev.gameres.com/Program/Abstract/Geometry.htm#
下面是代码

#include<iostream>

using namespace std;

struct node

{

    double x1,y1,x2,y2;

};

double cross(double x1,double y1,double x2,double y2)

{

    return x1*y2-x2*y1;

}

int main()

{

    int n;

    node p[110];

    while(cin>>n&&n!=0)

    {

       int i,j,ans=0;

       for(i=0;i<n;i++)

           cin>>p[i].x1>>p[i].y1>>p[i].x2>>p[i].y2;

       for(i=0;i<n-1;i++)

        for(j=i+1;j<n;j++)

       {

           if((cross(p[j].x1-p[i].x1,p[j].y1-p[i].y1,p[j].x2-p[j].x1,p[j].y2-p[j].y1))*cross(p[j].x1-p[i].x2,p[j].y1-p[i].y2,p[j].x2-p[j].x1,p[j].y2-p[j].y1)<=0)

            {

                if((cross(p[j].x1-p[i].x1,p[j].y1-p[i].y1,p[i].x2-p[i].x1,p[i].y2-p[i].y1))*cross(p[j].x2-p[i].x1,p[j].y2-p[i].y1,p[i].x2-p[i].x1,p[i].y2-p[i].y1)<=0)

                    ans++;

            }

       }

       cout<<ans<<endl;

    }

    return 0;

}