题目来源：

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): 6563    Accepted Submission(s): 3167


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
 

 分析：

完全是模板

代码如下：

#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <iostream>
#include <vector>
using namespace std;
typedef long long ll;
double EPS=1e-10;
// 考虑误差的加法运算
double add(double a,double b)
{
    if(fabs(a+b)<EPS*(fabs(a)+fabs(b))) return 0;
    return a+b;
}
struct Point{
    double x,y;
    Point(){}
    Point(double x,double y):x(x),y(y){} // 构造函数，方便代码编写
    Point operator +(Point p){
        return Point(add(x,p.x), add(y,p.y));
    }
    Point operator-(Point p){
        return Point(add(x,-p.x),add(y,-p.y));
    }
    Point operator*(double d){
        return Point(x*d,y*d);
    }
    double dot(Point p){  // 内积 点乘
        return add(x*p.x, y*p.y);
    }
    double xmult(Point p){//  外积 叉乘
        return add(x*p.y,-y*p.x);
    }
};
//判断点p0是否在线段p1p2内
int on_segment(Point p1, Point p2, Point p0)
{
    if (((p1-p0).x * (p2-p0).x <=0 )&& ((p1-p0).y * (p2-p0).y <=0))   // 中间是 &&
        return 1;
    return 0;
}
// 判断线段是否相交
int intersection(Point p1,Point p2, Point q1,Point q2)
{
    double d1=(p2-p1).xmult(q1-p1);           // 计算p1p2 到q 点的转向 d1>0 左转，  d1 <0 右转
    double d2=(p2-p1).xmult(q2-p1);
    double d3=(q2-q1).xmult(p1-q1);
    double d4=(q2-q1).xmult(p2-q1);
    if((d1==0 && on_segment(p1,p2,q1) )|| (d2==0 && on_segment(p1,p2,q2) )||
       (d3==0&& on_segment(q1,q2,p1)) || (d4==0 && on_segment(q1,q2,p2)))
       return 1;
    else if(d1*d2<0 && d3*d4 <0)    // 中间是 && 
        return 1;
    return 0;
}
int main() {
    int t,sum;
    Point p1[105],p2[105];
    while(cin>>t&&t)
    {
        sum=0;
        for(int i=0;i<t;i++)
        {
            cin>>p1[i].x>>p1[i].y>>p2[i].x>>p2[i].y;
        }
        for(int i=0;i<t;i++)
            for(int j=i+1;j<t;j++)
            sum+=intersection(p1[i],p2[i],p1[j],p2[j]);
        cout<<sum<<endl;
    }
    return 0;
}