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;
}