BZOJ 1094 ZJOI2007 粒子运动 计算几何

标签：bzoj   bzoj1094   计算几何   

题目大意：给定一个圆，一堆粒子在里面反射，每个粒子只能撞墙k次，求全程粒子间距离的最小值

每两个粒子之间计算一遍

反射就是把射线沿着切线作镜像变换

随便搞搞咯……

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define M 110
#define EPS 1e-7
#define INF 1e9
using namespace std;
typedef long double ld;
struct Point{
	ld x,y;
	Point() {}
	Point(ld _,ld __):
		x(_),y(__) {}
	friend istream& operator >> (istream &_,Point &p)
	{
		_>>p.x>>p.y;
		return _;
	}
	friend Point operator + (const Point &p1,const Point &p2)
	{
		return Point(p1.x+p2.x,p1.y+p2.y);
	}
	friend Point operator - (const Point &p1,const Point &p2)
	{
		return Point(p1.x-p2.x,p1.y-p2.y);
	}
	friend ld operator * (const Point &p1,const Point &p2)
	{
		return p1.x*p2.y-p1.y*p2.x;
	}
	friend Point operator * (const Point &p,ld rate)
	{
		return Point(p.x*rate,p.y*rate);
	}
	friend ld Distance(const Point &p1,const Point &p2)
	{
		return sqrt( (p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y) );
	}
}O;
struct Line{
	Point p,v;
	Line() {}
	Line(const Point &_,const Point &__):
		p(_),v(__) {}
	friend istream& operator >> (istream &_,Line &l)
	{
		cin>>l.p>>l.v;
		return _;
	}
	friend Point Get_Intersection(const Line &l1,const Line &l2)//计算两条直线的交点
	{
		Point u=l1.p-l2.p;
		ld temp=(l2.v*u)/(l1.v*l2.v);
		return l1.p+l1.v*temp;
	}
}a[M][M];
int n,k;
ld r,ans=INF;
ld t[M][M];
ld Quadratic_Equation(ld a,ld b,ld c)//解一元二次方程
{
	ld re1=( -b+sqrt(b*b-4*a*c) )/2/a;
	ld re2=( -b-sqrt(b*b-4*a*c) )/2/a;
	return max(re1,re2);
}
ld Get_Intersection(const Line &l)//计算直线与圆的交点
{
	return Quadratic_Equation(	l.v.x*l.v.x+l.v.y*l.v.y,
								2*(l.p.x*l.v.x+l.p.y*l.v.y),
								l.p.x*l.p.x+l.p.y*l.p.y-r*r 	);
}
Point Mirror(const Point &p,const Line &l)//求点p关于直线l的镜像
{
	Point u=Point(l.v.y,-l.v.x);
	Point intersection=Get_Intersection(Line(p,u),l);
	return intersection+(intersection-p);
}
ld Min_Distance(const Line &l1,const Line &l2,ld l,ld r)
{
	ld a=(l1.v.x-l2.v.x)*(l1.v.x-l2.v.x)+(l1.v.y-l2.v.y)*(l1.v.y-l2.v.y);
	ld b=(l1.p.x-l2.p.x)*(l1.v.x-l2.v.x)+(l1.p.y-l2.p.y)*(l1.v.y-l2.v.y);
	if(a<EPS) return Distance(l1.p+l1.v*r,l2.p+l2.v*r);
	ld t=-b/a;
	if(t>r) t=r;
	if(t<l) t=l;
	return Distance(l1.p+l1.v*t,l2.p+l2.v*t);
}
void Calculate(int x,int y)
{
	int i=1,j=1;
	ld last=0;
	while(i<=k&&j<=k)
	{
		if(t[x][i]>t[y][j])
			swap(x,y),swap(i,j);
		ans=min(ans,Min_Distance(a[x][i],a[y][j],last,t[x][i]) );
		last=t[x][i];i++;
	}
}
int main()
{
	int i,j;
	cin>>O>>r>>n>>k;
	for(i=1;i<=n;i++)
	{
		cin>>a[i][1];
		a[i][1].p=a[i][1].p-O;
		for(j=2;j<=k;j++)
		{
			t[i][j-1]=Get_Intersection(a[i][j-1]);
			Point intersection=a[i][j-1].p+a[i][j-1].v*t[i][j-1];
			Point intersection_T=Point(intersection.y,-intersection.x);
			a[i][j]=Line( 	Mirror(a[i][j-1].p,Line(intersection,intersection_T)) ,
							Mirror(a[i][j-1].v,Line(Point(0,0),intersection_T)) ) ;
		}
		t[i][k]=Get_Intersection(a[i][k]);
	}
	for(i=1;i<=n;i++)
		for(j=i+1;j<=n;j++)
			Calculate(i,j);
	cout<<fixed<<setprecision(3)<<ans<<endl;
	return 0;
}

BZOJ 1094 ZJOI2007 粒子运动 计算几何

标签：bzoj   bzoj1094   计算几何   

原文地址：http://blog.csdn.net/popoqqq/article/details/44980283