poj 3348 Cows

求凸包面积。

三角形面积等于叉积的一半~

/*
Coder:Zhaofa Fang
Lang:C++
*/
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#define pb push_back
using namespace std;

//==========================================
const double PI=acos(-1);

struct Point
{
    double x,y;
}point[10005],res[10005];

double det(double x1,double y1,double x2,double y2)
{
    return x1*y2-x2*y1;
}
double xmult(Point o,Point a ,Point b)
{
    return det(a.x-o.x,a.y-o.y,b.x-o.x,b.y-o.y);
}
double dis(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
bool cmp(const Point &x,const Point &y)
{
    if(xmult(point[0],x,y)==0)
        return dis(point[0],x)<dis(point[0],y);
    return xmult(point[0],x,y)>0;
}
double Graham(int n)
{
    Point tmp;
    int k=0,top=2;
    for(int i=1;i<n;i++)
    {
        if(point[i].y<point[k].y || point[i].y==point[k].y && point[i].x<point[k].x)
        k=i;
    }
    tmp=point[0],point[0]=point[k],point[k]=tmp;
    sort(point+1,point+n,cmp);
    res[0]=point[0],res[1]=point[1],res[2]=point[2];
    for(int i=3;i<n;i++)
    {
        while(top && xmult(res[top-1],res[top],point[i])<=0)top--;
        res[++top]=point[i];
    }
    double sum=0;
    for(int i=2;i<top+1;i++)
    {
        sum += xmult(res[0],res[i-1],res[i]);
    }
    return sum;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&point[i].x,&point[i].y);
        }
        printf("%d\n",(int)(Graham(n)/100));
    }
    return 0;
}