（叉积，线段判交）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): 12959    Accepted Submission(s): 6373

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

 

Sample Output

1

3

叉积求线段判交的参考链接：

https://www.cnblogs.com/Duahanlang/archive/2013/05/11/3073434.html

https://www.cnblogs.com/tuyang1129/p/9390376.html

C++代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
struct Point{
    double x1,y1,x2,y2;
}node[];
int cross(const Point &a, const Point &b){
    double k1 = (a.x2 - a.x1) * (b.y1 - a.y1) - (a.y2 - a.y1) * (b.x1 - a.x1);
    double k2 = (a.x2 - a.x1) * (b.y2 - a.y1) - (a.y2 - a.y1) * (b.x2 - a.x1);
    if(k1 * k2 <= ){
        return ;
    }
    else
        return ;
}
int main(){
    int n;
    while(scanf("%d",&n),n){
        int ans = ;
        for(int i = ; i < n; i++){
            scanf("%lf%lf%lf%lf",&node[i].x1,&node[i].y1,&node[i].x2,&node[i].y2);
        }
        for(int i = ; i < n-; i++){
            for(int j = i + ; j < n; j++){
                ans += (cross(node[i],node[j])) && (cross(node[j],node[i]));
            }
        }
        printf("%d\n",ans);
    }
    return ;
}