POJ 2074 | 线段相交

最新推荐文章于 2022-01-17 09:37:30 发布

转载最新推荐文章于 2022-01-17 09:37:30 发布 · 60 阅读

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/mrsheep/p/8011271.html

本文介绍了一个计算特定条件下房屋可见视野宽度的程序。该程序通过处理输入的房屋、道路及障碍物坐标来确定房屋沿道路方向的最大视野宽度。涉及到点运算、直线相交等几何计算方法。

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#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define eps 1e-8
using namespace std;
bool dcmp(double x,double y)
{
    if (fabs(x-y)>eps) return 1;
    return 0;
}
struct point
{
    double x,y;
    point () {};
    point (double _x,double _y)
	{
	    x=_x,y=_y;
	}
    point operator + (const point &a)const
	{
	    return point(x+a.x,y+a.y);
	}
    point operator - (const point &a) const
	{
	    return point(x-a.x,y-a.y);
	}
    double operator * (const point &a)const
	{
	    return x*a.y-a.x*y;
	}
    double dot (const point &a,const point &b)
	{
	    return a.x*b.x+a.y*b.y;
	}
    bool operator < (const point &a)const
	{
	    return x<a.x;
	}
    bool operator == (const point &a)const
	{
	    return dcmp(x,a.x) && dcmp(y,a.y);
	}
};
struct range
{
    double x1,x2,y;
    bool operator < (const range &a)const
	{
	    return x1<a.x1;
	}
}house,block[1010],road;
bool cmp(range a,range b)
{
    return a.x1<b.x1;
}
double findinter(point p1,point p2,point p3,point p4){    
    double ans=((p1.x-p3.x)*(p3.y-p4.y)-(p1.y-p3.y)*(p3.x-p4.x))/((p1.x-p2.x)*(p3.y-p4.y)-(p1.y-p2.y)*(p3.x-p4.x));    
   return  p1.x+(p2.x-p1.x)*ans;
}    
double getIntersect(point a,point b,point c,point d)
{
    double A1=b.y-a.y,B1=a.x-b.x,C1=(b.x-a.x)*a.y-(b.y-a.y)*a.x;
    double A2=d.y-c.y,B2=c.x-d.x,C2=(d.x-c.x)*c.y-(d.y-c.y)*c.x;
    return (C2*B1-C1*B2)/(A1*B2-A2*B1);
}
int n;
int main()
{
//    freopen("1.out","w",stdout);
    while (scanf("%lf%lf%lf",&house.x1,&house.x2,&house.y)!=0)
    {
	if (house.x1==0 && house.x2==0 && house.y==0) break;
	scanf("%lf%lf%lf",&road.x1,&road.x2,&road.y);
	scanf("%d",&n);
	for (int i=0;i<n;i++)
	    scanf("%lf%lf%lf",&block[i].x1,&block[i].x2,&block[i].y);
	sort(block,block+n,cmp);
	point p1,p2,p3,p4;
	p3=point(road.x1,road.y);
	p4=point(road.x2,road.y);
	double ans=-1,Lmax=-1.0;
	for (int i=0;i<=n;i++)
	{
	    double l,r;
	      if (block[i].y>=house.y) continue;
	    if (i==0)
		l=road.x1;
	    else
	    {
		p1=point(house.x1,house.y);
		p2=point(block[i-1].x2,block[i-1].y);
		l=findinter(p1,p2,p3,p4);
	    }
	    if (i==n)
		r=road.x2;
	    else
	    {
		p1=point(house.x2,house.y);
		p2=point(block[i].x1,block[i].y);
		r=findinter(p1,p2,p3,p4);
	    }
	    if (l<road.x1) l=road.x1;
	    if (r>road.x2) r=road.x2;
	    if (l<Lmax) l=Lmax;
	    Lmax=max(Lmax,l);
	    ans=max(ans,r-l);
	}
	if (ans<=0) puts("No View");
	else printf("%.2f\n",ans);
    }
    return 0;
}

转载于:https://www.cnblogs.com/mrsheep/p/8011271.html