Surround the Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7203    Accepted Submission(s): 2752

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.

   
   

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

   
   

    
    243.06
   
   

水平序的Andrew算法：

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

struct node
{
	double x,y;
}a[105],b[105];

double cmp(node n,node m)    //先比較X坐标，在比較Y坐标（从小到大）
{
	if(n.x != m.x)   
		return  n.x < m.x;
	else
		return  n.y < m.y;
}

double Cross(node a,node b,node c)     //计算叉积大小
{
	return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);
}

double dis(node a,node b)     //计算距离
{
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

int CH(node* a,int n,node* b)
{
	sort(a,a+n,cmp);
	int m=0,i;
	for(i=0;i<n;i++)    //从左往右，选下边界
	{
		while(m > 1 && Cross(b[m-2],b[m-1],a[i]) < 0)
			m--;
		b[m++]=a[i];
	}


	int k=m;
	for(i=n-2;i>=0;i--)    //从右往左，选上边界
	{
		while(m > k && Cross(b[m-2],b[m-1],a[i]) < 0)
			m--;
		b[m++]=a[i];
	}
	
	if(n >1)  m--;
	return m;
}

int main()
{
	int n;
	while(cin>>n)
	{
		if(n==0)   break;
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));

		int i,j;
		for(i=0;i<n;i++)
		{
			cin>>a[i].x>>a[i].y;
		}
		
	//	cout<<CH(a,n,b)<<endl;     //输出所选点的总数
		if(n==1)   
			cout<<0.00<<endl;
		else  if(n==2)
			printf("%.2lf\n",dis(a[0],a[1]));	
		else
		{
			int m=CH(a,n,b);
			double s=0;
			for(i=1;i<m;i++)
				s+=dis(b[i-1],b[i]);
			s+=dis(b[0],b[m-1]);
			printf("%.2lf\n",s);
		}
	//	for(i=0;i<CH(a,n,b);i++)    //输出所选点的坐标
	//		cout<<b[i].x<<" "<<b[i].y<<endl;   

	}

	return 0;
}