hdoj1086(计算线段有无焦点)

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本文介绍了一个几何问题：给定N条线段，计算所有交点的数量，并考虑多条线段在同一交点的情况。提供了完整的C++代码实现，使用矢量叉积判断线段是否相交。

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1976 Accepted Submission(s): 914

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
#include <iostream>
using namespace std;

double direction(double x1,double y1,double x2,double y2)//矢量叉积
{ return x1*y2-x2*y1; }

bool crossed(double A_x1,double A_y1,double A_x2,double A_y2,//判断两矢量相互之间的顺逆时针关系
    double B_x1,double B_y1,double B_x2,double B_y2)            
{
return (direction(A_x1-B_x1, A_y1-B_y1 , B_x2-B_x1, B_y2-B_y1)
   *direction(B_x2-B_x1, B_y2-B_y1, A_x2-B_x1, A_y2-B_y1) >= 0);       
}

struct SECMENT
{
double x1,y1;
double x2,y2;
}sec[100];

int main()
{
int N;
while(scanf("%d",&N) && N)
{
    for(int i=0;i<N;i++)
    {
        scanf("%lf %lf %lf %lf",&sec[i].x1, &sec[i].y1, &sec[i].x2, &sec[i].y2);
    }
    int cnt = 0;
    for(int i=0;i<N;i++)
    {
        for(int j=i+1;j<N;j++)//相交则需互相跨立
        {
          if(crossed(sec[i].x1, sec[i].y1, sec[i].x2, sec[i].y2, 
                     sec[j].x1, sec[j].y1, sec[j].x2, sec[j].y2)
    && crossed(sec[j].x1, sec[j].y1, sec[j].x2, sec[j].y2, 
                     sec[i].x1, sec[i].y1, sec[i].x2, sec[i].y2) )
            cnt++;
        }
    }
    printf("%d/n",cnt);
}
return true;
}