POJ - 3348 Cows 凸包求面积

题目链接：https://cn.vjudge.net/problem/POJ-3348

题意：给出n个点，求所形成的区域能放多少cow，一个cow占50

题解：先求下凸包上的点，在叉积求下面积即可

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
#define eps 1e-8
struct Point{
	double x, y;
	Point (double x_=0,double y_=0)
	{
		x=x_;y=y_;
	}
	Point operator + (const Point &xx)const 
	{
		return Point(x+xx.x,y+xx.y);
	}
	Point operator - (const Point &xx)const 
	{
		return Point(x-xx.x,y-xx.y);
	}
	
	
}a[10010],b[10010];
double cross(Point xx,Point yy)
{
	return (xx.x*yy.y-xx.y*yy.x);
}
double dis(Point xx,Point yy)
{
	return sqrt((xx.x-yy.x)*(xx.x-yy.x) + (xx.y-yy.y)*(xx.y-yy.y));
}
int n,m;
bool cmp(Point xx,Point yy)
{
	double O=cross(xx-a[1],yy-a[1]);
	if(O>0 || O==0&&dis(a[1],xx)<dis(a[1],yy)) return 1;
	else return 0;
}
void solve()
{
	for(int i=1;i<=min(2,n);i++)
		b[++m]=a[i];
	for(int i=3;i<=n;i++)
	{
		while(m>1 && cross(b[m]-b[m-1],a[i]-b[m-1])<=0)m--;
		b[++m]=a[i];
	}
	double ans=0;
	for(int i=2;i<m;i++)
	{
		ans+=cross(b[i]-b[1],b[i+1]-b[1])/2;
	}
//	cout<<ans<<endl;
	printf("%d\n",(int)ans/50);
}
int main()
{
	while(~scanf("%d",&n))
	{
		int k=1;
		m=0;
		for(int i=1;i<=n;i++)
		{
			scanf("%lf%lf",&a[i].x,&a[i].y);
			if(a[i].y<a[k].y|| (a[i].y==a[k].y&&a[i].x<a[k].x))
			{
				k=i;
			}
		}
		swap(a[1].x,a[k].x);swap(a[1].y,a[k].y);
		sort(a+2,a+1+n,cmp);
		solve();
	}
	return 0;
}