Uva 11168 Airport（凸包）

题目地址：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2109

思路：

1.所找直线一定为凸包上的边，故可枚举凸包上的边，求最短距离和。

2.由于所有点在直线同侧所以所有abs(A*x+B*y+C)/sqrt(A*A+B*B)等于(A*x+B*y+C)/sqrt(A*A+B*B)或-(A*x+B*y+C)/sqrt(A*A+B*B)。则点到直线的距离和可以O(1)时间内获得：fabs(A*sum(x[i])+B*sum(y[i])+C*(n-2)) (i=1---n且i不在所枚举的直线上)

3.两点式转化为一般式：











4.若只有一个点，结果为0.

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define debug
using namespace std;
const double eps=1e-6;
struct Point
{
double x,y;
Point(double x=0.0,double y=0.0):x(x),y(y) {}
void read()
{
scanf("%lf%lf",&x,&y);
}
};
typedef Point Vector;

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

bool operator < (const Point &a,const Point &b)
{
return a.x<b.x||(a.x==b.x&&a.y<b.y);
}

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

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

int ConvexHull(Point* p,int n,Point* ch)
{
sort(p,p+n);
int m=0;
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--;
return m;
}

const int maxn=10000+50;
const double INF=1e20;

int n;
Point p[maxn],convex[maxn];

int main()
{
#ifdef debu
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif // debug
int t,cas=0;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
double ans=INF;
double totx=0,toty=0;
for(int i=0; i<n; i++)
{
p[i].read();
totx+=p[i].x;
toty+=p[i].y;
}
if(n==1)
{
printf("Case #%d: %.3f\n",++cas,0.0);
continue;
}
int num=ConvexHull(p,n,convex);
for(int i=0; i<num; i++)
{
double A,B,C,tot=0;
Point p1=convex[i],p2=convex[(i+1)%num];
A=p2.y-p1.y,B=p1.x-p2.x,C=p2.x*p1.y-p1.x*p2.y;
tot+=(totx-p1.x-p2.x)*A+(toty-p1.y-p2.y)*B+C*(n-2);
tot/=sqrt(A*A+B*B);
ans=min(ans,fabs(tot));
}
printf("Case #%d: %.3f\n",++cas,ans/n);
}
return 0;
}