HDU 1086 You can Solve a Geometry Problem too

A

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1086

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

 

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<iostream>#define eps 1e-8#define zero(x) (((x)>0?(x):-(x))<eps)using namespace std;struct point{    double x,y;};struct line{    point a,b;} q[10001];//计算cross product (P1 - P0) * (P2 - P0) double xmult(point p1,point p2,point p0){    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}//判断点是否在线段上，包括端点int dot_online_in(point p,line l){    return zero(xmult(p,l.a,l.b))&&(l.a.x-p.x)*(l.b.x-p.x)<eps&&(l.a.y-p.y)*(l.b.y-p.y)<eps;}//判断三点共线int dots_inline(point p1,point p2,point p3){    return zero(xmult(p1,p2,p3));}//判断两点在线段同侧，点在线段上返回0int same_side(point p1,point p2,line l){    return xmult(l.a,p1,l.b)*xmult(l.a,p2,l.b)>eps;}//判断两线段相交，包括端点和部分重合！！！int intersect_in(line u,line v){    if (!dots_inline(u.a,u.b,v.a)||!dots_inline(u.a,u.b,v.b))    {        return !same_side(u.a,u.b,v)&&!same_side(v.a,v.b,u);    }    return dot_online_in(u.a,v)||dot_online_in(u.b,v)||dot_online_in(v.a,u)||dot_online_in(v.b,u);}int main(){    int n,m;    int i,j;    while(scanf("%d",&n)!=EOF)    {        if(n == 0)        {            break;        }        int count = 0;        for(i=0; i<n; i++)        {            scanf("%lf%lf%lf%lf",&q[i].a.x,&q[i].a.y,&q[i].b.x,&q[i].b.y);        }        for(i=0; i<n; i++)        {            for(j=i+1; j<n; j++)            {                int pp = 0;                pp = intersect_in(q[i],q[j]);                if(pp == 1)                {                    count++;                }            }        }        printf("%d\n",count);    }    return 0;}