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本文介绍了一个算法问题，即如何利用给定的树木位置坐标来构建最大的牧场，并计算该牧场能够容纳多少头牛。通过计算几何的方法，寻找这些点构成的最大凸包并计算其面积。

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题目：

Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are forced to save money on buying fence posts by using trees as fence posts wherever possible. Given the locations of some trees, you are to help farmers try to create the largest pasture that is possible. Not all the trees will need to be used.

However, because you will oversee the construction of the pasture yourself, all the farmers want to know is how many cows they can put in the pasture. It is well known that a cow needs at least 50 square metres of pasture to survive.

Input：

The first line of input contains a single integer, n (1 ≤ n ≤ 10000), containing the number of trees that grow on the available land. The next n lines contain the integer coordinates of each tree given as two integers x and y separated by one space (where -1000 ≤ x, y ≤ 1000). The integer coordinates correlate exactly to distance in metres (e.g., the distance between coordinate (10; 11) and (11; 11) is one metre).

Output：

You are to output a single integer value, the number of cows that can survive on the largest field you can construct using the available trees.

Sample Input：

Sample Output：

PS：

本题的大概意思就是给你一些点，求出其凸包，算出其面积。主要理解其算他的凸包。。。。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define N 10010
struct Point
{  
    double x,y;         
    Point(long double _ = .0,long double __ = .0):x(_),y(__) {}  
    friend bool operator <(const Point &a,const Point &b)
   {return a.x<b.x||(a.x==b.y&&a.y<b.y);}
 // Point operator +(const Point &a)const 
   //{return Point(x+a.x,y+a.y);}  
    Point operator -(const Point &a)const 
    {return Point(x-a.x,y-a.y);}  
   // Point operator *(double a)const 
   // {return Point(x*a,y*a);}
}p[N],ch[N];

///int sign(double a)
//{return(a>eps)?1:(a<-eps)?-1:0;}
//double dot(Point a,Point b)
//{return a.x*b.x+a.y*b.y;}
double cross(Point a,Point b)
{return a.x*b.y-a.y*b.x;}

int ComvexHull(Point* p,int n,Point* ch)
{
    sort(p+1,p+n+1);
    int m=0;
    for(int i=1;i<=n;++i)
    {
        while(m>1&&cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
        ch[m++]=p[i];//下凸包 
    }
    int k=m;
    for(int i=n-1;i>=1;--i)
    {
        while(m>k&&cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
        ch[m++]=p[i];
    }//上凸包 
    if(n>1) m--;
    return m;
}
double PolygonArea(Point* p,int n)
{
    double area=0;
    for(int i=2;i<=n-1;++i)
        area+=cross(p[i]-p[1],p[i+1]-p[1]);
    return area/2;
}
int n;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;++i) scanf("%lf%lf",&p[i].x,&p[i].y);
    int m=ComvexHull(p,n,ch);
    printf("%d\n",(int)PolygonArea(ch,m)/50);
    return 0;
}