HDU 1086: 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): 6997 accepted submission (s): 3385


Problem descriptionmany ry (ry) problems were designed in the ACM/ICPC. and now, I also prepare a geometry problem for this final exam. according to the experience of specified 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 (line segment), please output the number of all intersections (intersection ). you shoshould count repeatedly if M (M> 2) segments intersect at the same point.

Note:
You can assume that two segments wocould not intersect at more than one point.
 
Inputinput 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.
 
Outputfor each case, print the number of intersections, and one line one case.
 
Sample Input

20.00 0.00 1.00 1.000.00 1.00 1.00 0.0030.00 0.00 1.00 1.000.00 1.00 1.00 0.0000.00 0.00 1.00 0.000

 
Sample output

13 This is a geometric question, that is, whether you can determine whether two straight lines have cross points. The rest is easy. Determine whether there is an intersection between the line AB and the CD: both conditions are met: ('x' indicates the Cross Product) 1. point C and point D are on both sides of AB. (vector (abxac) * (abxad) <= 0) 2. point A and point B are on both sides of the CD. (vector (cdxca) * (cdxcb) <= 0)
#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<cmath>using namespace std;struct Node{    double x1, y1, x2, y2;}point[105];int n;double work(double x1, double y1, double x2, double y2){    return x1 * y2 - x2 * y1;}bool judge(int i, int j){    double a = work(point[i].x1 - point[j].x1, point[i].y1 - point[j].y1, point[i].x1 - point[i].x2, point[i].y1 - point[i].y2);    double b = work(point[i].x1 - point[i].x2, point[i].y1 - point[i].y2, point[i].x1 - point[j].x2, point[i].y1 - point[j].y2);    a = a * b;    double c = work(point[j].x1 - point[i].x1, point[j].y1 - point[i].y1, point[j].x1 - point[j].x2, point[j].y1 - point[j].y2);    double d = work(point[j].x1 - point[j].x2, point[j].y1 - point[j].y2, point[j].x1 - point[i].x2, point[j].y1 - point[i].y2);    c = c * d;    if(a >= 0 && c >= 0)        return true;    return false;}int main(){    while(cin >> n, n){        for(int i = 0; i < n; i++)            cin >> point[i].x1 >> point[i].y1 >> point[i].x2 >> point[i].y2;        int count = 0;        for(int i = 0; i < n; i++)        {            for(int j = i + 1; j < n; j++)                if(judge(i, j))                    count++;        }        cout << count << endl; }    return 0;}