hdoj-1086-You can Solve a Geometry Problem too 判断线段是否相交

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8683 Accepted Submission(s): 4227

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

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

Author

lcy

Recommend

We have carefully selected several similar problems for you:1392 2108 2150 1348 1147

#include<stdio.h>#include<math.h>#include<algorithm>using namespace std;const double esp=1e-7;struct point{double x,y;}p[110],q[110];double cross(point a,point b,point c){return (c.x-a.x)*(c.y-b.y)-(c.x-b.x)*(c.y-a.y);}bool insert(point aa,point bb,point cc,point dd){if(max(aa.x,bb.x)<min(cc.x,dd.x))  return false;if(max(aa.y,bb.y)<min(cc.y,dd.y))  return false;if(max(cc.x,dd.x)<min(aa.x,bb.x))  return false;if(max(cc.y,dd.y)<min(aa.y,bb.y))  return false;if(cross(aa,bb,cc)*cross(aa,bb,dd)>0) return false;if(cross(cc,dd,aa)*cross(cc,dd,bb)>0) return false;return true;}int main(){int n;while(~scanf("%d",&n),n){int i,j;double s1,s2;for(i=1;i<=n;++i){scanf("%lf%lf%lf%lf",&p[i].x,&p[i].y,&q[i].x,&q[i].y);}int ncase=0;for(i=1;i<=n;++i){for(j=i+1;j<=n;++j){    if(insert(p[i],q[i],p[j],q[j])) ncase++;}}printf("%d\n",ncase);     }return 0;}