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HDU|HDU1086_You can Solve a Geometry Problem too(几何/叉积判断点线关系)

解题报告
题目传送门
题意：
求线段有多少相交的。
【HDU|HDU1086_You can Solve a Geometry Problem too(几何/叉积判断点线关系)】思路：
AB，CD两线段相交，
也就是C，D两点在线段AB的两端。A，B两点在线段CD的两端。
用叉积判断点在线段的左右。

#include #include #include using namespace std; struct Point { double x,y; }; struct Seg { Point l,r; } L[110]; double xmulti(Point a,Point b,Point p) { return (b.x-a.x)*(p.y-a.y)-(p.x-a.x)*(b.y-a.y); } int main() { int n,cnt,i,j; while(cin>>n&&n) { cnt=0; memset(L,0,sizeof(L)); for(i=0; i


You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7347Accepted Submission(s): 3573


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


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