UVA11168 Airport

题意

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分析

首先发现距离最短的直线肯定在凸包上面。

然后考虑直线一般方程\(Ax+By+C=0\)，点\((x_0,y_0)\)到该直线的距离为

\[\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}
\]

由于所有点在直线同侧，所以绝对值里面的符号相同，所以维护所有点\(x\)坐标和\(y\)坐标之和就行了。

时间复杂度\(O(T n \log n)\)

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<ctime>
#include<cstring>
#define rg register
#define il inline
#define co const
template<class T>il T read()
{
	rg T data=0;
	rg int w=1;
	rg char ch=getchar();
	while(!isdigit(ch))
	{
		if(ch=='-')
			w=-1;
		ch=getchar();
	}
	while(isdigit(ch))
	{
		data=data*10+ch-'0';
		ch=getchar();
	}
	return data*w;
}
template<class T>T read(T&x)
{
	return x=read<T>();
}
using namespace std;
typedef long long ll;

struct Point
{
	double x,y;

	Point(double x=0,double y=0)
	:x(x),y(y){}

	bool operator<(co Point&rhs)
	{
		return x<rhs.x||(x==rhs.x&&y<rhs.y);
	}

	bool operator==(co Point&rhs)
	{
		return x==rhs.x&&y==rhs.y;
	}
};
typedef Point Vector;

Vector operator-(co Point&A,co Point&B)
{
	return Vector(A.x-B.x,A.y-B.y);
}

double Cross(co Vector&A,co Vector&B)
{
	return A.x*B.y-A.y*B.x;
}

vector<Point> ConvexHull(vector<Point> p)
{
	sort(p.begin(),p.end());
	p.erase(unique(p.begin(),p.end()),p.end());

	int n=p.size();
	int m=0;
	vector<Point>ch(n+1);
	for(int i=0;i<n;++i)
	{
		while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)
			--m;
		ch[m++]=p[i];
	}
	int k=m;
	for(int i=n-2;i>=0;--i)
	{
		while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)
			--m;
		ch[m++]=p[i];
	}
	if(n>1)
		--m;
	ch.resize(m);
	return ch;
}

// (x2-x1)(y-y1) = (y2-y1)(x-x1) -> ax+by+c=0
void LineGeneralEquation(co Point&p1,co Point&p2,double&a,double&b,double&c)
{
	a=p2.y-p1.y;
	b=p1.x-p2.x;
	c=-a*p1.x-b*p1.y;
}

int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	int T=read<int>();
	for(int kase=1;kase<=T;++kase)
	{
		int n=read<int>();
		vector<Point>P;
		double sumx=0,sumy=0;
		for(int i=0;i<n;++i)
		{
			int x=read<int>(),y=read<int>();
			sumx+=x,sumy+=y;
			P.push_back(Point(x,y));
		}
		vector<Point>ch=ConvexHull(P);
		int m=ch.size();
		double ans=1e9;
		if(m<=2)
			ans=0;
		else
			for(int i=0;i<m;++i)
			{
				double a,b,c;
				LineGeneralEquation(ch[i],ch[(i+1)%m],a,b,c);
				ans=min(ans,fabs(a*sumx+b*sumy+c*n)/sqrt(a*a+b*b));
			}
		printf("Case #%d: %.3lf\n",kase,ans/n);
	}
	return 0;
}