POJ 1228 (stable convex hull)

Language: Grandpa's EstateTime Limit: 1000 MS Memory Limit: 10000 KTotal Submissions: 8990 Accepted: 2383 DescriptionKamran the Believer inherits a grandmother's convex polygon estate. the perimeter of the manor is surrounded by ropes and wooden piles. but some ropes and stakes are missing. check whether the manor surrounded by remaining wooden piles is a stable convex bag (that is, the remaining nail can determine a unique convex bag ). the number of data groups in the first row of Input www.2cto.com is t (1 <= t <= 10 ). for each group of data, the first behavior n (1 <= n <= 1000) indicates the number of wooden piles. in the next n rows, the coordinates (x, y) of each behavior are fixed to an integer. output outputs YES or NO for each group of data, indicating whether the data is stable. sample Input16 0 01 23 42 02 4 5 0 Sample Outpu TNOSourceTehran 2002 Preliminary to ensure stability of a convex hull, if and only if any side of the convex hull has more than three wooden piles (including endpoints) proof: After the convex hull is created, enumerate, click 3rd on the edge. [Cpp] # include <cstdio> # include <cstring> # include <cstdlib> # include <cmath> # include <cctype> # include <iostream> # include <functional> # include <algorithm> using namespace std; # define MAXT (10 + 10) # define MAXN (1000 + 10) // # define sqr (x) (x * x) int sqr (int x) {return x * x;} struct P {int x, y; P () {} P (int _ x, int _ y): x (_ x ), y (_ y) {} friend istream & operator> (istream & cin, P & a) {cin>. x>. y; return cin;} Friend double dis (P a, P B) {return sqrt (double (sqr (. x-b.x) + sqr (. y-b.y);} a [MAXN]; struct V {int x, y; V () {} V (int _ x, int _ y ): x (_ x), y (_ y) {} V (P a, P B): x (B. x-a.x), y. y-a.y) {} friend int operator * (V a, V B) {return. x * B. y-a.y * B. x ;}}; int cmp (p a, p B) {int temp = V (a [1], A) * V (a [1], B ); if (temp> 0) return 1; else if (temp = 0 & dis (a [1], A) <dis (a [1], B )) return 1; else return 0;} int t, n, st [MAXN ]; Bool solve () {int size = 1; st [0] = 1; st [1] = 1; int j = 2; while (j <= n) {if (size <2 | V (a [st [size-1], a [st [size]) * V (a [st [size], a [j])> 0) {st [++ size] = j ++;} else size --;} a [++ n] = a [1]; st [+ size] = n; for (int I = 1; I <size; I ++) {/* int k = st [I-1] + 1; (; k <st [I + 1]; k ++) if (k! = St [I] & V (a [st [I], a [st [I + 1]) * V (a [st [I], a [st [k]) = 0) break; if (k = st [I + 1]) return 0; */int k = 1; (; k <n; k ++) if (k! = St [I] & k! = St [I + 1] & V (a [st [I], a [st [I + 1]) * V (a [st [I], a [k]) = 0) break; if (k = n) return 0;} return size> = 4;} int main () {// freopen ("poj1228.in", "r", stdin); cin> t; while (t --) {cin> n; for (int I = 1; I <= n; I ++) cin> a [I]; if (n <6) {cout <"NO \ n"; continue ;} int p = 1; for (int I = 2; I <= n; I ++) if (a [I]. x <a [p]. x | (a [I]. x = a [p]. x) & (a [I]. y <a [p]. y) p = I; swap (a [1], a [p]); sort (a + 2, a + 1 + n, cmp ); // cout <dis (P (), P () <dis (P (), P (); if (solve ()) cout <"YES \ n"; else cout <"NO \ n";} return 0;} remarks: If you replace sqr with define, "nan" is output"