POJ 3348 Cows

#include<cmath>
#include<cstdio>
#include<algorithm>
#define db double
using namespace std;
const int N=1e4+50;
const db eps=1e-9;
struct Point
{
    db x,y;Point(){}
    Point(db _x,db _y) {x=_x,y=_y;}
}p[N],q[N];
Point operator - (Point A,Point B) {return Point(A.x-B.x,A.y-B.y);} 
db operator * (Point A,Point B) {return A.x*B.x+A.y*B.y;} 
db operator ^ (Point A,Point B) {return A.x*B.y-A.y*B.x;} 
db dist(Point a,Point b) {return sqrt((a-b)*(a-b));} 
bool cmp(Point a,Point b)
{
    db c=(a-p[1])^(b-p[1]);
    if(fabs(c)<=eps) return dist(a,p[1])<dist(b,p[1]);
    return c<eps;
}
int main()
{
    int n;while(scanf("%d",&n)!=EOF)
    {
        if(n==1||n==2) {puts("0");continue;}
        for(int i=1;i<=n;i++) 
        {
            scanf("%lf%lf",&p[i].x,&p[i].y);
            if(i>1&&p[i].y<p[1].y) swap(p[i],p[1]);
        }
        sort(p+2,p+n+1,cmp); int ed=1; q[ed]=p[1];
        for(int i=2;i<=n;i++)
        {
            while(ed>1&&((p[i]-q[ed-1])^(q[ed]-q[ed-1]))<eps) ed--;
            q[++ed]=p[i];
        } db res=0;
        for(int i=2;i<ed;i++) res+=((q[i+1]-q[1])^(q[i]-q[1]))/2;
// 有向面积的一半就是该三角形的面积
        printf("%d\n",(int)(res/50));
    }
    return 0;
}