凸包

Graham求凸包 O(N * logN)

/*==================================================*\
| Graham求凸包 O(N * logN)
| CALL: nr = graham(pnt, int n, res);
|pnt[]为给定点集 n为点的个数 res[]为凸包点集;
\*==================================================*/
struct Tpoint {int x,y;}g[MM]; //注意点的类型
Tpoint res[MM];
bool mult(Tpoint sp, Tpoint ep, Tpoint op){
    return (sp.x-op.x) * (ep.y-op.y)>=(ep.x-op.x)*(sp.y-op.y);
}
bool operator<(const Tpoint &l, const Tpoint &r){
    return l.y<r.y || (l.y==r.y && l.x<r.x);
}
int graham(Tpoint pnt[], int64 n, Tpoint res[]){
    int i, len, k = 0, 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
}