Language: Default Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 24604   Accepted: 8183 Description King想在自己的n个城堡外建Wall,使Wall与任一城堡距离至少为L且能围住它的城堡.  求Wall最短长度.  Input 第一行为 N (3 <= N <= 1000) 和 L(1 <= L <= 1000). 接下来 N 行,每行为城堡坐标Xi,Yi (-10000 <= Xi, Yi <= 10000). Output 输出一行Wall最短长度. Sample Input 9 100 200 400 300 400 300 300 400 300 400 400 500 400 500 200 350 200 200 200 Sample Output 1628 Hint 结果四舍五入就可以了 Source Northeastern Europe 2001

易证Wall最短长度=对城堡求凸包的周长+2πL

这里介绍一个凸包求法QuickHull

备注Quickhull-

s表示向量CA与CB的叉积(在上一求得，用于求距离），初始化为0

用[l,r]表示点集区间，

有2个要点

1.A，B , C的关系

(1)

(2)

2.i,j的调换

可从下图看出

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<functional>
#include<algorithm>
#include<cctype>
using namespace std;
#define MAXN (1000+10)
#define MAXXi (10000+10)
#define eps (1e-10)
const double pi=3.1415926;
int sqr(int x) {return x*x;}
struct P
{
	int x,y;
	P(){}
	P(int _x,int _y):x(_x),y(_y){}
	friend istream& operator>>(istream &cin,P &a)
	{
		cin>>a.x>>a.y;
		return cin;
	}
	friend bool operator<(const P a,const P b){return (a.x==b.x)?a.y<b.y:a.x<b.x;}
	friend bool operator>(const P a,const P b){return (a.x==b.x)?a.y>b.y:a.x>b.x;}
}a[MAXN],st[MAXN];
double dis(P A,P B)
{
	return sqrt((double)sqr(A.x-B.x)+sqr(A.y-B.y));
}
struct V
{
	int x,y;
	V(){}
	V(int _x,int _y):x(_x),y(_y){}
	V(P a,P b):x(b.x-a.x),y(b.y-a.y){}
	friend int operator*(V a,V b)
	{
		return a.x*b.y-a.y*b.x;
	}
};
int n,l,size=0;
double s[MAXN]={0.0};
void hull(int l,int r,P A,P B)
{
	int p=l;
	for (int k=l;k<=r;k++)
		if (s[p]<s[k]||s[p]==s[k]&&a[p]<a[k]) p=k;
	P C=a[p];
	int i=l-1,j=r+1;
	for (int k=l;k<=r;k++)
	{
		s[++i]=V(a[k],A)*V(a[k],C);
		if (s[i]>0) /*Right*/ swap(a[i],a[k]); else i--;
	}
	for (int k=r;k>=l;k--)
	{
		s[--j]=V(a[k],C)*V(a[k],B);
		if (s[j]>0) /*Right*/ swap(a[j],a[k]); else j++;
	}
	if (l<=i) hull(l,i,A,C);
	st[++size]=C;
	if (j<=r) hull(j,r,C,B);
}
int main()
{
//	freopen("poj1113.in","r",stdin);
	cin>>n>>l;
	for (int i=1;i<=n;i++)
		cin>>a[i];
	int pmin=1,pmax=1;
	for (int i=2;i<=n;i++)
	{
		if (a[i]<a[pmin]) pmin=i;
		if (a[i]>a[pmax]) pmax=i;
	}
	swap(a[1],a[pmin]);swap(a[n],a[pmax]);
	st[++size]=a[1];
	hull(2,n,a[1],a[1]);
//	for (int i=1;i<=size;i++) cout<<st[i].x<<' '<<st[i].y<<' '<<endl;
	double ans=0;
	for (int i=1;i<=size;i++) ans+=dis(st[i],st[i+1<=size?i+1:1]);
	cout.setf(ios::fixed);
	cout.precision(0);
	cout<<ans+2*pi*l<<endl;
	return 0;
}