HDU 1086 You can Solve a Geometry Problem too（判断线段相交）

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本文介绍了一个简单的几何问题：计算给定线段集合中所有交点的数量，并允许在相同点处的多次相交计入总数。通过解析输入数据并利用向量乘积判断线段是否相交来实现。

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12006 Accepted Submission(s): 5946

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

可以重复计数，就是说把每个线段与其余线段比较，相交就+1.

#include<string.h>
#include<stdio.h>
using namespace std;
struct point{
	double x,y;
};
struct node
{
	point s,e;
}mat[105];
double product(point &a,point &b,point &c)
{
	double x1,y1,x2,y2;
	x1=a.x-c.x;
	y1=a.y-c.y;
	x2=b.x-c.x;
	y2=b.y-c.y;
	return (x1*y2-x2*y1);
}
int main()
{
	int n;
	while(~scanf("%d",&n)&&n)
	{
		node s1,t1;
		for(int i=0;i<n;i++)
		scanf("%lf%lf%lf%lf",&mat[i].s.x,&mat[i].s.y,&mat[i].e.x,&mat[i].e.y);
		int sum=0;
		for(int i=0;i<n-1;i++)
			for(int j=i+1;j<n;j++)
		if(product(mat[i].s,mat[i].e,mat[j].s)*product(mat[i].s,mat[i].e,mat[j].e)<=0&&product(mat[j].s,mat[j].e,mat[i].s)*product(mat[j].s,mat[j].e,mat[i].e)<=0)
		{//相交 
			sum++;
		}
		printf("%d\n",sum);
	}
	return 0;
}