hdu 1086（计算几何入门题——计算线段交点个数）

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1086

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7167    Accepted Submission(s): 3480

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

//////////////////////////////////////////////////////////////////////////////直接上的模板解决的，不知是喜是忧啊

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <iostream>
#include <algorithm>

using namespace std;
const int MAX=;
const double eps = 1e-;

struct point
{
    double x,y;
};

struct beline
{
    point a,b;
};

point p[MAX];
int n=;
bool dy(double x,double y)
{
  return x>y+eps;
}
bool xy(double x,double y)
{
    return x<y-eps;
}
bool dyd(double x,double y)
{
    return x > y - eps;
}
bool xyd(double x,double y)
{
    return x<y+eps;
}
bool dd(double x,double y)
{
    return fabs(x-y) < eps;
}
double crossProduct(point a,point b,point c)
{
    return (c.x - a.x)*(b.y-a.y)-(b.x-a.x)*(c.y-a.y);
}
bool onSegment(point a,point b,point c)
{
    double maxx=max(a.x,b.x);
    double maxy=max(a.y,b.y);
    double minx=min(a.x,b.x);
    double miny=min(a.y,b.y);
    if(dd(crossProduct(a,b,c),0.0)&&dyd(c.x,minx)&&xyd(c.x,maxx)&&dyd(c.y,miny)&&xyd(c.y,maxy))
        return true;
    return false;
}

bool segIntersect(point p1,point p2,point p3,point p4)
{
    double d1 = crossProduct(p3,p4,p1);
    double d2 = crossProduct(p3,p4,p2);
    double d3 = crossProduct(p1,p2,p3);
    double d4 = crossProduct(p1,p2,p4);
    if(xy(d1 * d2,0.0)&&xy(d3*d4,0.0))
        return true;
    if(dd(d1,0.0)&&onSegment(p3,p4,p1))
        return true;
    if(dd(d2,0.0)&&onSegment(p3,p4,p2))
        return true;
    if(dd(d3,0.0)&&onSegment(p1,p2,p3))
        return true;
    if(dd(d4,0.0)&&onSegment(p1,p2,p4))
        return true;
    return false;
}

int main()
{
    int cas,i,j;
    while(scanf("%d",&cas)!=EOF&&cas!=)
    {
         beline L[MAX];
         n=;
         for(i=;i<cas;i++)
         {
             scanf("%lf%lf%lf%lf",&L[i].a.x,&L[i].a.y,&L[i].b.x,&L[i].b.y);
         }
         for(i=;i<cas;i++)
         {
             for(j=i+;j<cas;j++)
                 if(segIntersect(L[i].a,L[i].b,L[j].a,L[j].b))
                 {
                     n++;
                 }
         }
         printf("%d\n",n);
    }
    return ;
}