Graham求凸包

Graham

/*
*  Graham 求凸包 O(N * logN)
*  CALL: nr = graham(pnt, int n, res); res[]为凸包点集;
*/
struct point
{
double x, y;
};

bool mult(point sp, point ep, point op)
{
return (sp.x - op.x) * (ep.y - op.y) >= (ep.x - op.x) * (sp.y - op.y);
}

//inline bool operator < (const point &l, const point &r)
//{
//    return l.y < r.y || (l.y == r.y && l.x < r.x);
//}

int graham(point pnt[], int n, point res[])
{
int i, len, top = 1;
sort(pnt, pnt + n);
if (n == 0)
{
return 0;
}
res[0] = pnt[0];
if (n == 1)
{
return 1;
}
res[1] = pnt[1];
if (n == 2)
{
return 2;
}
res[2] = pnt[2];
for (i = 2; i < n; i++)
{
while (top && mult(pnt[i], res[top], res[top - 1]))
{
top--;
}
res[++top] = pnt[i];
}
len = top;
res[++top] = pnt[n - 2];
for (i = n - 3; i >= 0; i--)
{
while (top != len && mult(pnt[i], res[top], res[top - 1]))
{
top--;
}
res[++top] = pnt[i];
}
return top; //  返回凸包中点的个数
}