Cows - POJ 3348(凸包求面积)

题目大意：利用n棵树当木桩修建牛圈，知道每头牛需要50平的生存空间，求最多能放养多少头牛。

分析：赤裸裸的求凸包然后计算凸包的面积。

代码如下：

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#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<math.h>
using namespace std;

const int MAXN = 1e4+;
const double EPS = 1e-;

int sta[MAXN], top;

struct point
{
    double x, y;

    point(double x=, double y=):x(x), y(y){}
    point operator - (const point &t)const{
        return point(x-t.x, y-t.y);
    }
    double operator *(const point &t)const{
        return x*t.x + y*t.y;
    }
    double operator ^(const point &t)const{
        return x*t.y - y*t.x;
    }
}p[MAXN];
int Sign(double t)
{
    if(t > EPS)return ;
    if(fabs(t) < EPS)return ;
    return -;
}
double Dist(point a, point b)
{
    return sqrt((a-b)*(a-b));
}
bool cmp(point a, point b)
{
    int t = Sign((a-p[])^(b-p[]));

    if(t == )
        return Dist(a, p[]) < Dist(b, p[]);
    return t > ;
}
void Graham(int N)
{
    int k=;

    for(int i=; i<N; i++)
    {
        if(p[k].y>p[i].y || (Sign(p[k].y-p[i].y)== && p[k].x > p[i].x))
            k = i;
    }
    swap(p[], p[k]);
    sort(p+, p+N, cmp);

    sta[]=, sta[]=, top=;

    if(N < )
    {
        top = N-;
        return ;
    }

    for(int i=; i<N; i++)
    {
        while(top> && Sign((p[i]-p[sta[top]])^(p[sta[top-]]-p[sta[top]])) <= )
            top--;
        sta[++top] = i;
    }
}
int main()
{
    int N;

    while(scanf("%d", &N) != EOF && N)
    {
        for(int i=; i<N; i++)
        {
            scanf("%lf%lf", &p[i].x, &p[i].y);
        }

        Graham(N);

        double area=;

        for(int i=; i<top; i++)
        {
            double a = Dist( p[ sta[] ], p[ sta[i] ] );
            double b = Dist( p[ sta[] ], p[ sta[i+] ] );
            double c = Dist( p[ sta[i] ], p[ sta[i+] ] );
            double q = (a+b+c) / ;

            area += sqrt(q*(q-a)*(q-b)*(q-c));
        }

        printf("%d\n", (int)(area/50.0));
    }

    return ;
}