Seeking convex hull (I was too dishes)

Today, school brother asked to write a little about it ... it seems to be called (Graham scan method)?
(Only this one) too dishes

In fact, seeking convex hull is very simple (that is, Why did not learn? Is cry myself vegetables)
must first find the lower left corner. (It must be a point on the convex hull)
at that point as the center of the polar angle sorting
and then drained order the first two points on the stack
cross product of vectors determined stack configuration penultimate point and the last point and the current point of determination, if the element is negative, de-stacked, has retreated negative or not the stack tree less than 2
put the stack is point determination

Until all points traversed over, the stack element is the point on the convex hull ...

Code is as follows :( really nothing worth more to say ...)

#include <iostream>
#include <algorithm>
#define MAX 1000
using namespace std;

struct node {
    int x, y;

    node(int _x = 0, int _y = 0)
    {
        x = _x;
        y = _y;
    }

    node operator -(node a)//向量减
    {
        return node(x - a.x, y - a.y);
    }

    int operator*(node a)//叉乘
    {
        return (x * a.y - y * a.x);
    }

    friend int getdis(node a, node b)//两点间距离的平方
    {
        return ((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
    }

    void show()
    {
        cout << "x:" << x << " y:" << y << endl;
    }

}p[MAX];

int ans[MAX];
int n;

int cmp(node a, node b)//极角排序
{
    int t = (a - p[0]) * (b - p[0]);
    if (t != 0)
        return t > 0;
    else
        return getdis(a, p[0]) < getdis(b, p[0]);
}

void getans()
{
    int len = 0;
    ans[len++] = 0;
    for (int i = 1; i < n; i++)
    {
        while (len >= 2 && ((p[ans[len - 1]] - p[ans[len - 2]]) * (p[i] - p[ans[len - 2]]) <= 0)) len--;
        ans[len++] = i;
    }//求凸包...
    for (int i = 0; i < len; i++)
        p[ans[i]].show();
    int s = 0;
    for (int i = 1; i+1 < len; i++)
    {
        s += (p[ans[i]] - p[0]) * (p[ans[i + 1]] - p[0]);
    }//求凸多边形面积(因为我这里都是整数,就没用double...正常写肯定要用double的...)
    cout << 1.0 * s / 2;
}


int main()
{
    cin >> n;
    for (int i = 0; i < n; i++)
        cin >> p[i].x >> p[i].y;
    int m = 0;//最左下的点的下标
    for (int i = 1; i < n; i++)
    {
        if (p[i].y < p[m].y)
            m = i;
        else if (p[i].y == p[m].y && p[i].x < p[m].x)
            m = i;
    }
    node t = p[m];
    p[m] = p[0];
    p[0] = t;//一定要把最左下的点和最开始的点换位置!!这里处理不好还会出莫名其妙的BUG(不知道为什么,对sort研究不深...)
    sort(p, p + n, cmp);
    getans();
    return 0;
}


/*
9
1 0
3 2
2 0
2 1
2 2
2 3
1 1
0 0
3 3
*/