Grandpa's estate --- poj1228 (convex hull)

Http://poj.org/problem? Id = 1228

Students say this is a stable convex bag. I think it is a convex bag.

The so-called stability is to determine whether the original convex hull can be added to get a larger convex hull, and this convex hull contains all the points on the original convex hull. Knowing this is simple. After finding the convex hull, you can determine whether the points on each edge are more than three points. ------ Don't worry.

The question cannot be understood. It means that each side of the convex hull has at least three points.

In addition, if there is only one line, no is output.

# Include <stdio. h> # include <string. h> # include <stdlib. h> # include <math. h> # include <algorithm> # include <iostream> # include <queue> using namespace STD; # define n 1005 # define PI ACOs (-1.0) # define ESP 1e-8int s [N]; struct point {Double X, Y; point (double x = 0, Double Y = 0): x (x), y (y) {} point operator-(const point & temp) const {return point (x-temp.x, y-temp.y);} point operator + (const point & temp) const {return point (x + temp. x, Y + temp. y);} bool operator = (const point & temp) const {return (x = temp. X & Y = temp. y);} double operator * (const point & temp) const {return (x * temp. X + Y * temp. y);} int operator ^ (const point & temp) const {double T = x * temp. y-y * temp. x; If (T> ESP) return 1; if (FABS (t) <ESP) return 0; Return-1 ;}} P [N]; double dist (point, point B) {return SQRT (a-B) * (a-B);} int CMP (Point A1, point A2) {int t = (a1-p [0]) ^ (a2-p [0]); If (t = 0) return dist (A1, P [0]) <dist (A2, P [0]); else return T> 0 ;}int top; void Graham (int n) // returns the convex hull {s [0] = 0; s [1] = 1; Top = 1; for (INT I = 2; I <n; I ++) {While (top> 0 & (P [I]-P [s [Top]) ^ (P [s [Top-1]-P [s [Top]) <= 0) Top --; s [++ top] = I ;}} int main () {int t, n; scanf ("% d", & T); While (t --) {int K = 0; scanf ("% d ", & N); For (INT I = 0; I <n; I ++) {scanf ("% lf", & P [I]. x, & P [I]. y); If (P [I]. Y <p [K]. Y | (P [I]. y = P [K]. Y & P [I]. x <p [K]. x) k = I;} swap (P [0], p [k]); sort (p + 1, P + N, CMP); Graham (N ); s [++ top] = s [0]; int flag = 0; For (INT I = 1; I <= top; I ++) {int ans = 0; for (Int J = 0; j <n; j ++) {If (P [J]-P [s [I]) ^ (P [s [I-1]-P [s [I]) = 0) ans ++;} If (ANS <3) {flag = 1; break ;}} if (flag ==0 & top> = 3) printf ("Yes \ n"); else printf ("NO \ n ");} return 0 ;}

 

Grandpa's estate --- poj1228 (convex hull)