Poj 1228 Grandpa's estate (a portion of vertices on a given convex hull determine whether the convex hull is unique)

Question link: http://poj.org/problem? Id = 1228

This question is hard to think about. At first, it was always wa. Later, I didn't know where to change it.

From the example, we can see that the points of the convex Hill do not come in order, but are random points.

Then we create a convex hull on these points, so we cannot remove the edges when creating a convex hull.

That is, even the points that can be removed cannot be removed, that is, the traditional graphm method cannot be pre-selected.

If we want this convex hull to be unique after the end, we cannot add any point to the outside to make it grow.

A larger convex hull requires a vertex to have at least two vertices before or after it.

The edges can be collocated. Otherwise, you can insert a convex hull to the vertex!

# Include <iostream> # include <stdio. h> # include <string. h >#include <algorithm> # include <cmath> using namespace STD; # define EPS 1e-6 # define PI 3.14159265 struct point {Double X; Double Y;} po [1500], temp; int N, Pos; bool zero (double A) {return FABS (a) <EPS;} double DIS (point & A, point & B) // returns the square {return (. x-b.x) * (. x-b.x) +. y-b.y) * (. y-b.y);} double round SS (point & A, point & B, point & C) // calculate the X product of a B and A c {r Eturn (B. x-a.x) * (C. y-a.y)-(B. y-a.y) * (C. x-a.x);} int CMP (const void * a, const void * B) {int X; X = pair SS (po [0], * (point *), * (point *) B); If (zero (x) return DIS (po [0], * (point *) a) <DIS (po [0], * (point *) B )? -1: 1; return second SS (po [0], * (point *) A, * (point *) B)> 1e-8? -1: 1;} int Graham (INT num) {bool flag = true; int I, j, k = 2; po [num] = po [0]; // fangbiannum ++; for (I = 3; I <num; I ++) {While (Primary SS (po [k-1], po [K], po [I]) <-EPS) {k --;} po [++ K] = po [I]; // This loop ends, no more !} N = K; k = 0; for (I = 0; I <n; I ++) if (zero (distinct SS (po [I], po [I + 1], po [I + 2]) {k = 1; continue;} else {flag = false; If (k = 0) {printf ("NO \ n "); k = 10; return 0;} k = 0;} If (FLAG) {printf ("NO \ n"); Return 0;} If (K! = 10) printf ("Yes \ n"); Return 0;} int main () {int I, J, K, T; point my_temp, temp1; scanf ("% d", & T); While (t --) {scanf ("% d", & N); scanf ("% lf ", & po [0]. x, & po [0]. y); temp = po [0]; Pos = 0; for (I = 1; I <n; I ++) {scanf ("% lf ", & po [I]. x, & po [I]. y); If (po [I]. Y <temp. y) temp = po [I], Pos = I; If (po [I]. y = temp. Y & po [I]. x> temp. x) {temp1 = temp; temp = po [I], Pos = I;} temp = temp1; If (n <6) {printf ("NO \ n "); continue;} my_temp = po [0]; po [0] = po [POS]; po [POS] = my_temp; qsort (PO + 1, n-1, sizeof (po [0]), CMP); Graham (n);} return 0 ;}