BZOJ 1043 [HAOI2008]下落的圆盘 ——计算几何

腊鸡题目。

我们考虑倒着加入圆盘,考虑每一个圆被之后覆盖的弧的总长度。

首先需要把包含的情况去掉。

然后用解析几何算出相交的弧的位置。

然后求一个交。

然后把剩下的结果加入答案中。

写了一下午,发现cos值可能大于1,不能用acos函数计算。

QAQ

#include <map>
#include <cmath>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define F(i,j,k) for (int i=j;i<=k;++i)
#define D(i,j,k) for (int i=j;i>=k;--i)
#define eps 1e-8
#define sqr(x) x*x
#define ll long long
#define mp make_pair

const double pi=acos(-1.0);

struct Vector{
	double x,y;
	void print()
	{
		printf("Vector -> (%.3f,%.3f)\n",x,y);
	}
};

struct Point{
	double x,y;
	void print()
	{
		printf("Point . (%.3f,%.3f)\n",x,y);
	}
};

Vector operator + (Vector a,Vector b)
{Vector ret;ret.x=a.x+b.x;ret.y=a.y+b.y;return ret;}

Vector operator - (Vector a,Vector b)
{Vector ret;ret.x=a.x-b.x;ret.y=a.y-b.y;return ret;}

Point operator + (Point a,Vector b)
{Point ret;ret.x=a.x+b.x;ret.y=a.y+b.y;return ret;}

Vector operator - (Point a,Point b)
{Vector ret;ret.x=a.x-b.x;ret.y=a.y-b.y;return ret;}

Vector operator * (Vector a,double b)
{Vector ret;ret.x=a.x*b;ret.y=a.y*b;return ret;}

double operator * (Vector a,Vector b)
{return a.x*b.y-a.y*b.x;}

Vector Turn(Vector a,double b)
{
	Vector ret;
	ret.x=a.x*cos(b)-a.y*sin(b);
	ret.y=a.x*sin(b)+a.y*cos(b);
	return ret;
}

int n;

struct Circle{
	double x,y,r;
	void print()
	{
		printf("Circle (%.3f,%.3f) r = %.3f\n",x,y,r);
	}
}a[1005];

double dst(double x1,double y1,double x2,double y2)
{return sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));}

int dcmp(double x)
{
	if (x>eps) return 1;
	if (fabs(x)<eps) return 0;
	return -1;
}

bool in(Circle a,Circle b)
{
	if (dcmp(dst(a.x,a.y,b.x,b.y)-(b.r-a.r))<=0) return true;
	return false;
}

struct Cut_Cir{
	double l,r;
	void print()
	{
		printf("Ang (%.3f,%.3f)\n",l,r);
	}
}sta[1000005];
int top=0;

double nege(double x)
{
	if (dcmp(x)==-1) x+=2*pi;
	while (dcmp(x-2*pi)>=0) x-=2*pi;
	return x;
}

void Cut(int pos)
{
	top=0; double dist; Vector v0,v1,v2;
	F(i,pos+1,n)
		if ((dist=dst(a[pos].x,a[pos].y,a[i].x,a[i].y))<a[pos].r+a[i].r)
		{
			double ang; ang=(sqr(a[pos].r)+sqr(dist)-sqr(a[i].r))/(2*a[pos].r*dist);
			if (ang>1.0) continue;
			ang=acos(ang);
			v0.x=a[i].x-a[pos].x;v0.y=a[i].y-a[pos].y;
			v0=v0*(1.0/sqrt(sqr(v0.x)+sqr(v0.y))*a[pos].r);
			v1=Turn(v0,ang);
			v2=Turn(v0,-ang);
			double t1=atan2(v1.y,v1.x),t2=atan2(v2.y,v2.x);
			t1+=pi;t2+=pi;
			while (t1>2*pi) t1-=2*pi;
			while (t2>2*pi) t2-=2*pi;
			if (t1<t2)
			{
				++top;
				sta[top].l=0;
				sta[top].r=t1;
				++top;
				sta[top].l=t2;
				sta[top].r=2*pi;
			}
			else
			{
				++top;
				sta[top].l=t2;
				sta[top].r=t1;
			}
		}
}

double C(Circle a)
{return a.r*2*pi;}

bool cmp2(Cut_Cir a,Cut_Cir b)
{return a.l<b.l;}

double cal(double r)
{
	sort(sta+1,sta+top+1,cmp2);
	double now=0,ret=0;
	F(i,1,top)
	{
		now=max(now,sta[i].l);
		ret+=max(0.0,sta[i].r-now);
		now=max(now,sta[i].r);
	}
	return ret*r;
}

double ans=0;

int main()
{
	scanf("%d",&n);
	F(i,1,n) scanf("%lf%lf%lf",&a[i].r,&a[i].x,&a[i].y);
	F(i,1,n)
	{
		int flag=1;
		F(j,i+1,n) if (in(a[i],a[j]))
			flag=0;
		if (!flag) continue;
		Cut(i);
		ans+=C(a[i]);
		ans-=cal(a[i].r);
	}
	printf("%.3f\n",ans);
}

  

posted @ 2017-04-06 17:03  SfailSth  阅读(96)  评论(0编辑  收藏  举报