【计算几何初步-线段相交】【HDU1089】线段交点

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7847    Accepted Submission(s): 3834

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

 

求线段相交分为两个讨论

1.规范相交

2.不规范相交

对于1 直接跨立实验叉积搞定

对与2 在跨立实验后 发现有三点共线情况 利用点积判断第二个向量的那个点是否落在第一个向量中间

几点注意

1.写一个doublesgn函数  来避免-0.00001当做小于0的情况

这是根据上述定义写的丑陋代码

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313
#define exp 10e-6
using namespace std;
struct point
{
	double x;
	double y;
};
point start[101],end[101];
int N,ans=0;
void init()
{
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
}
int dblcmp(double d)
{
	if(fabs(d)<exp) return 0;
	else return (d>0)?1:-1;
}
void input()
{
	ans=0;
	for(int i=1;i<=N;i++)
	{
		scanf("%lf%lf%lf%lf",&start[i].x,&start[i].y,&end[i].x,&end[i].y);
	}
}
double XX(double x1,double y1,double x2,double y2)
{
	return x1*y2-x2*y1;
}
int cross(point &a,point &b,point &c)
{
	return dblcmp(XX(b.x-a.x,b.y-a.y,c.x-a.x,c.y-a.y));
}
double DX(double x1,double y1,double x2,double y2)
{
	return x1*x2+y1*y2;
}
int Dcross(point &a,point &b,point &c)
{
	return dblcmp(DX(b.x-a.x,b.y-a.y,c.x-a.x,c.y-a.y));
}
int panX(point &S1,point &E1,point &S2,point &E2)
{
	int SS2=cross(S1,E1,S2),EE2=cross(S1,E1,E2),SS1=cross(S2,E2,S1),EE1=cross(S2,E2,E1);
	int a1=SS2*EE2;
	int a2=SS1*EE1;
	if(a1==1||a2==1) return 0;
	if(a1==-1&&a2==-1) return 1;
	if(a1==0)
	{
		if(SS2==0)
		{
			if(Dcross(S1,E1,S2)>=0&&Dcross(E1,S1,S2)>=0) return 1;
			else return 0;
		}
		if(EE2==0)
		{
			if(Dcross(S1,E1,E2)>=0&&Dcross(E1,S1,E2)>=0) return 1;
			else return 0;
		}
	}
	if(a2==0)
	{
		if(SS1==0)
		{
			if(Dcross(S2,E2,S1)>=0&&Dcross(E2,S2,S1)>=0) return 1;
			else return 0;
		}
		if(EE1==0)
		{
			if(Dcross(S2,E2,E1)>=0&&Dcross(E2,S2,E1)>=0) return 1;
			else return 0;
		}
	}
	return 1;
}
void solve()
{
	for(int i=1;i<=N;i++)
	 for(int j=i+1;j<=N;j++)
	 {
	 	if(panX(start[i],end[i],start[j],end[j]))
	 	ans++;
	 }
}
int main()
{
	//	init();
	while(scanf("%d",&N)!=EOF&&N)
	{

		input();
		solve();
		cout<<ans<<endl;
	}
	return 0;
}
  

傻逼的发现 点积那里只要判断一次就好了 改了改

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313
#define exp 10e-6
using namespace std;
struct point
{
	double x;
	double y;
};
point start[101],end[101];
int N,ans=0;
void init()
{
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
}
int dblcmp(double d)
{
	if(fabs(d)<exp) return 0;
	else return (d>0)?1:-1;
}
void input()
{
	ans=0;
	for(int i=1;i<=N;i++)
	{
		scanf("%lf%lf%lf%lf",&start[i].x,&start[i].y,&end[i].x,&end[i].y);
	}
}
double XX(double x1,double y1,double x2,double y2)
{
	return x1*y2-x2*y1;
}
int cross(point &a,point &b,point &c)
{
	return dblcmp(XX(b.x-a.x,b.y-a.y,c.x-a.x,c.y-a.y));
}
double DX(double x1,double y1,double x2,double y2)
{
	return x1*x2+y1*y2;
}
int Dcross(point &a,point &b,point &c)
{
	return dblcmp(DX(b.x-a.x,b.y-a.y,c.x-a.x,c.y-a.y));
}
int panX(point &S1,point &E1,point &S2,point &E2)
{
	int SS2=cross(S1,E1,S2),EE2=cross(S1,E1,E2),SS1=cross(S2,E2,S1),EE1=cross(S2,E2,E1);
	int a1=SS2*EE2;
	int a2=SS1*EE1;
	if(a1==1||a2==1) return 0;
	if(a1==-1&&a2==-1) return 1;
	if(a1==0)
	{
		if(SS2==0)
		{
			if(Dcross(S2,E1,S1)<=0) return 1;
			else return 0;
		}
		if(EE2==0)
		{
			if(Dcross(E2,E1,S1)<=0) return 1;
			else return 0;
		}
	}
	if(a2==0)
	{
		if(SS1==0)
		{
			if(Dcross(S1,E2,S2)<=0) return 1;
			else return 0;
		}
		if(EE1==0)
		{
			if(Dcross(E1,E2,S2)<=0) return 1;
			else return 0;
		}
	}
	return 1;
}
void solve()
{
	for(int i=1;i<=N;i++)
	 for(int j=i+1;j<=N;j++)
	 {
	 	if(panX(start[i],end[i],start[j],end[j]))
	 	ans++;
	 }
}
int main()
{
	//	init();
	while(scanf("%d",&N)!=EOF&&N)
	{

		input();
		solve();
		cout<<ans<<endl;
	}
	return 0;
}

用胡浩大牛的模板再写一遍

#include <cstdio>
#include <cmath>
#include <algorithm>
#include<iostream>

using namespace std;

#define M 250
#define INF 0xFFFFFFF

const double eps = 1e-8;
const double inf = 10000;
const int maxP = 1100;
const double PI = acos(-1.0);

inline double sqr(double d)
{
	return d * d;
}

inline int sgn(double d)
{
	return d < -eps? -1: d > eps;
}

struct Point
{
	double x, y;

	Point(const double &_x = 0, const double &_y = 0) : x(_x), y(_y) {}

	bool operator == (const Point &p) const
	{
		return sgn(x - p.x) == 0 && sgn(y - p.y) == 0;
	}

	bool operator < (const Point &p) const
	{
		return y + eps < p.y || (y < p.y + eps && x + eps < p.x);
	}

	Point operator + (const Point &p) const
	{
		return Point(x + p.x, y + p.y);
	}

	Point operator - (const Point &p) const
	{
		return Point(x - p.x, y - p.y);
	}

	Point operator * (const double &k) const
	{
		return Point(x * k, y * k);
	}

	Point operator / (const double &k) const
	{
		return Point(x / k, y / k);
	}
	//叉积: <0：p在x的逆时针方向, =0: 共线, >0：顺时针方向
	double operator *(const Point &p) const
	{
		return x * p.y - y * p.x;
	}
	//点积：<0 :钝角， =0 :直角, >0 :锐角
	double operator / (const Point &p) const
	{
		return x * p.x + y * p.y;
	}

	double len2()
	{
		return x * x + y * y;
	}

	double len()
	{
		return sqrt(x * x + y * y);
	}
	//向量变化为对应单位向量的k倍
	Point scale(const double &k)
	{
		return sgn(len())? (*this) * (k / len()): (*this);
	}
	//点关于y = x对称， 再关于y轴对称（向量逆时针旋转90度）
	Point turnLeft()
	{
		return Point(-y, x);
	}
	//点关于y = x对称， 再关于x轴对称（向量顺时针旋转90度）
	Point turnRight()
	{
		return Point(y, -x);
	}

	void input()
	{
		scanf("%lf%lf", &x, &y);
	}

	void output()
	{
		printf("%.2lf %.2lf\n", x + eps, y + eps);
	}

	double Distance(Point p)
	{
		return sqrt(sqr(p.x - x) + sqr(p.y - y));
	}
	//以点P位轴逆时针转angle角度， 再放大k倍
	Point rotate(const Point &p, double angle, double k = 1)
	{
		Point vec = (*this) - p;
		double Cos(cos(angle) * k), Sin(sin(angle) * k);
		return p + Point(vec.x * Cos - vec.y * Sin, vec.x * Sin + vec.y * Cos);
	}
};

struct Line
{
	Point a , b;

	Line(const Point &_a = 0, const Point &_b = 0): a(_a), b(_b) {}

	Line(double c, double d, double e, double f): a(Point(c, d)), b(Point(e, f)) {}
	//<0： 点p在直线左边， =0: 点p在直线上， >0: 点p在直线右边
	double operator * (const Point &p) const
	{
		return (b - a) * (p - a);
	}
	//>0: 角apb为锐角， =0：角apb为直角， <0: 角apb为钝角
	double operator / (const Point &p) const
	{
		return (p - a) / (p - b);
	}

	void input()
	{
		a.input();
		b.input();
	}

	void output()
	{
		a.output();
		b.output();
	}

	double len()
	{
		return a.Distance(b);
	}
	//是否和直线v共线
	bool parallel(Line v)
	{
		return !sgn((b - a) * (v.b - v.a));
	}
	//是否和线段v相交， 交点是p
	bool SegCrossSeg(const Line &v, Point &p)
	{
		double s1 = v * a, s2 = v * b;
		if(sgn(s2 - s1) == 0)
			return false;
		p = (a * s2 - b * s1) / (s2 - s1);
		return (sgn(v / p) <= 0 && sgn((*this) / p) <= 0);
	}
	//线段与线段v 2: 规范相交， 1: 不规范相交, 0: 不相交
	int SegCrossSeg(const Line &v)
	{
		int d1 = sgn((*this) * v.a);
		int d2 = sgn((*this) * v.b);
		int d3 = sgn(v * a);
		int d4 = sgn(v * b);
		if((d1 ^ d2) == -2 && (d3 ^ d4) == -2)
			return 2;
		return ((d1 == 0 && sgn((*this) / v.a) <= 0)
				|| (d2 == 0 && sgn(( *this ) / v.b) <= 0)
				|| (d3 == 0 && sgn(v / a) <= 0)
				|| (d4 == 0 && sgn(v / b) <= 0));
	}
	//直线与线段v 2: 规范相交， 1: 不规范相交, 0: 不相交
	int LineCrossSeg(const Line &v)
	{
		int d1 = sgn((*this) * v.a) , d2 = sgn((*this) * v.b);
		if((d1 ^ d2) == -2)
			return 2;
		return (d1 == 0 || d2 == 0);
	}
	//直线与直线v 2: 规范相交， 1: 重合, 0: 不相交
	int LineCrossLine(const Line &v)
	{
		if((*this).parallel(v))
			return (sgn(v * a) == 0);
		return 2;
	}
	//返回两条直线的交点
	Point CrossPoint(const Line &v)
	{
		double s1 = v * a , s2 = v * b;
		return (a * s2 - b * s1) / (s2 - s1);
	}
	//返回点p到线段的最短距离
	double DisPointToSeg(Point p)
	{
		if(a == b)
			return a.Distance(p);
		Point q = p + (a - b).turnLeft();
		if(((p - a) * (q - a)) * ((p - b) * (q - b)) > 0)
			return min(p.Distance(a), p.Distance(b));
		return fabs((*this) * p) / a.Distance(b);
	}
	//返回点p到线段的最短距离的点
	Point PointToSeg(Point p)
	{
		if(a == b)
			return a;
		Point q = p + ( a - b ).turnLeft();
		if(((p - a) * (q - a)) * ((p - b) * (q - b)) > 0)
			return p.Distance(a) < p.Distance(b) ? a : b;
		return CrossPoint(Line(p, q));
	}
	//返回点P到直线的距离
	double DisPointToLine(const Point &p)
	{
		return fabs((*this) * p) / a.Distance(b);
	}
	//返回过点P与直线垂直的直线的交点
	Point PointToLine(const Point &p)
	{
		return CrossPoint(Line(p, p + (a - b).turnLeft()));
	}
	//返回点Q, 直线PQ平行该直线
	Point SymPoint(const Point &p)
	{
		return PointToLine(p) * 2 - p;
	}
	//向法线方向平移d
	void Move(const double &d)
	{
		Point t = (b - a).turnLeft().scale(d);
		a = a + t, b = b + t;
	}
};
Line A[101];
int N;
int ans;
void input()
{
	ans=0;
	for(int i=1;i<=N;i++)
	{
		scanf("%lf%lf%lf%lf",&A[i].a.x,&A[i].a.y,&A[i].b.x,&A[i].b.y);
	}
}
void solve()
{
	for(int i=1;i<=N;i++)
	 for(int j=i+1;j<=N;j++)
	 {
	 	if(A[i].SegCrossSeg(A[j]))
	 	ans++;
	 }
}
int main()
{
	while(scanf("%d",&N)!=EOF&&N)
	{

		input();
		solve();
		cout<<ans<<endl;
	}
	return 0;
}