HDU 1086 you can solve a geometry problem too (calculate the number of intersection points of a line segment template)

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1086

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You can solve a geometry problem too Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others) Total submission (s): 7227 accepted submission (s): 3504

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

Authorlcy
Apply the geida template;
The Code is as follows:

#include <cstdio>#include <iostream>#include <algorithm>using namespace std;const double eps=1e-10;struct point{ double x, y;};double min(double a, double b) { return a < b ? a : b;}double max(double a, double b) { return a > b ? a : b; }bool inter(point a, point b, point c, point d){if (min(a.x, b.x) > max(c.x, d.x)||min(a.y, b.y) > max(c.y, d.y)||min(c.x, d.x) > max(a.x, b.x)||min(c.y, d.y) > max(a.y, b.y) ) return 0;double h, i, j, k;h = (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);i = (b.x - a.x) * (d.y - a.y) - (b.y - a.y) * (d.x - a.x);j = (d.x - c.x) * (a.y - c.y) - (d.y - c.y) * (a.x - c.x);k = (d.x - c.x) * (b.y - c.y) - (d.y - c.y) * (b.x - c.x);return h * i <= eps && j * k <= eps;}int main(){int n;int i, j;point p[117][2];while(cin>>n && n){for(i = 0; i < n; i++){cin>>p[i][0].x>>p[i][0].y>>p[i][1].x>>p[i][1].y;}int count = 0;for(i = 0; i < n; i++){for(j = i+1; j < n; j++){if(inter(p[i][0],p[i][1],p[j][0],p[j][1]))count++;}}cout<<count<<endl;}return 0;}

HDU 1086 you can solve a geometry problem too (calculate the number of intersection points of a line segment template)