Surround the Trees&&凸包入门题

Problem Description

There are a lot of trees in an area. A peasant wants to buy a rope to surround all these trees. So at first he must know the minimal required length of the rope. However, he does not know how to calculate it. Can you help him?
The diameter and length of the trees are omitted, which means a tree can be seen as a point. The thickness of the rope is also omitted which means a rope can be seen as a line.



There are no more than 100 trees.

Input

The input contains one or more data sets. At first line of each input data set is number of trees in this data set, it is followed by series of coordinates of the trees. Each coordinate is a positive integer pair, and each integer is less than 32767. Each pair is separated by blank.

Zero at line for number of trees terminates the input for your program.

Output

The minimal length of the rope. The precision should be 10^-2.

Sample Input

   
   
   
   

    
    
    
    9 
12 7 
24 9 
30 5 
41 9 
80 7 
50 87 
22 9 
45 1 
50 7 
0 
   
   
   
   

Sample Output

   
   
   
   

    
    
    
    243.06
   
   
   
   
   
   
   
   

    
    
    
    AC代码：
   
   
   
   
   
   
   
   

    
    
    
    
#include<iostream>
#include<string.h>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#define N 105
using namespace std;
struct Point
{
	int  x;
	int  y;
}p[N];
int s[N],n,top;
int cross(const Point& a,const Point& b,const Point& c)
{return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);}
double dis(const Point& a,const Point& b)
{return sqrt((double)(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}
bool cmp(const Point& a,const Point& b)
{
	int res=cross(p[0],a,b);
	if(res>0||(res==0&&dis(p[0],a)<dis(p[0],b)))return true;
	return false;
}
void graham()
{
	s[0]=0;
	s[1]=1;
	top=1;
	for(int i=2;i!=n;++i)
	{
		while(top&&cross(p[s[top-1]],p[s[top]],p[i])<0) top--;
		 s[++top]=i;
	}
}
int main()
{
	while(~scanf("%d",&n),n)
	{
		for(int i=0;i!=n;++i)
			cin>>p[i].x>>p[i].y;
		cout.setf(ios::fixed);//设置cout为定点输出格式		cout.precision(2);//保留两位有效数字
		if(n==1) {
			cout<<"0.00"<<endl;
			continue;
		}
		else if(n==2) {
			cout<<dis(p[0],p[1])<<endl;
			continue;
		}
		Point ans=p[0];
		int u=0;
		for(int i=1;i!=n;++i)
			if(ans.y>p[i].y||(ans.y==p[i].y&&ans.x>p[i].x)) ans=p[i],u=i;
		  if(u) swap(p[u],p[0]);
		  sort(p+1,p+n,cmp);
		  graham();
		  double res=dis(p[s[0]],p[s[top]]);
		  for(int i=0;i!=top;++i)
			  res+=dis(p[s[i]],p[s[i+1]]);
		  cout<<res<<endl;
		 }return 0;
}