hdu1086 You can Solve a Geometry Problem too 计算几何求线段交点个数，快速排斥实验+跨立实验

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8229    Accepted Submission(s): 4007

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

 哎，，不得不说，数学不好的ACMer简直痛苦！！！

快速排斥实验和跨立实验

  看了一下午一道水题wrong到现在！！

上两个个好博客：http://www.cnblogs.com/g0feng/archive/2012/05/18/2508293.html

http://blog.sina.com.cn/s/blog_71dbfe2e0101f7zb.html

又get了一个新技能啊！！数学好是多么重要啊！！！

代码：

#include 
#include 
#define MAX 110
struct Segment{
	double x1,y1,x2,y2 ;
}seg[MAX];

double max(double a , double b)
{
	return a>b?a:b ;
}
double min(double a , double b)
{
	return amin(maxTX,maxRX) && max(minTY,minRY)>min(maxTY,maxRY))
	{
		return false ;
	}
	return true ;
}
double f(double x1 , double y1 ,double x2 , double y2)
{
	return x1*y2-y1*x2 ;
}
bool ifIntersect(Segment a, Segment b)
{
	if(quickExclude(a,b))
	{
		if(f(a.x1-b.x1,a.y1-b.y1,b.x2-b.x1,b.y2-b.y1)*f(b.x2-b.x1,b.y2-b.y1,a.x2-b.x1,a.y2-b.y1)>=0 &&
			f(b.x1-a.x1,b.y1-a.y1,a.x2-a.x1,a.y2-a.y1)*f(a.x2-a.x1,a.y2-a.y1,b.x2-a.x1,b.y2-a.y1)>=0)
			{
				return true ;
			}
	}
	return false ;
}
int main()
{
	int n ;
	while(~scanf("%d",&n) && n)
	{
		for(int i = 0 ; i < n ; ++i)
		{
			scanf("%lf%lf%lf%lf",&seg[i].x1,&seg[i].y1,&seg[i].x2,&seg[i].y2) ;
		}
		int ans = 0 ;
		for(int i = 0 ; i < n ; ++i)
		{
			for(int j = i+1 ; j < n ; ++j)
			{
				if(ifIntersect(seg[i],seg[j]))
				{
					++ans;
				}
			}
		}
		printf("%d\n",ans) ;
	}
	return 0 ;
}

与君共勉