[Convex Hull] HDU 3285

Because the regular convex hull is saved counter-clockwise, you only need to find the point in the upper left corner, and then go back to it to OK! The original template did not judge the common points, resulting in n times of Wa !!!

# Include <iostream> # include <vector> # include <map> # include <stack> # include <algorithm> # include <queue> # include <list> # include <set> # include <string. h> # include <stdlib. h> # include <math. h> # include <stdio. h> # include <ctype. h> # include <iomanip> using namespace STD; # define ll long # define PI ACOs (-1) # define fre freopen ("a.txt", "r", stdin) # define INF 9999999999 # define EPS 1e-6 # define N 55 struct point {int X, Y;}; Poin T p [N], chull [N], P0, stck [N]; int m; // Number of convex hull vertices int N; // All Points // (B,) X (C, A) int CNT; int cross (point a, point B, point C) {return (B. x-a.x) * (C. y-a.y)-(B. y-a.y) * (C. x-a.x);} // <0 then Ac in AB clockwise, need clockwise turn int DIS (point a, point B) {return (. x-b.x) * (. x-b.x) +. y-b.y) * (. y-b.y);} // sorting bool CMP (point a, point B) {int T = cross (P0, A, B ); return T> 0 | (t = 0 & DIS (P0, A) <DIS (P0, B )); // T> 0 indicates that p0b is in p0a counterclockwise} void convexhull () {int I, J, K; M = 0; CNT = 0; For (k = 0, I = 0; I <n; I ++) if (P [I]. Y <p [K]. Y | (P [I]. y = P [K]. Y & P [I]. x <p [K]. x) k = I; p0 = P [k]; // base point P [k] = P [0]; P [0] = P0; sort (p + 1, P + N, CMP); stck [0] = P [0]; stck [1] = P [1];/* int flag = 0; if (Cross (stck [0], stck [1], p [2]) = 0) {flag = 1; stck [1] = P [2];} if (FLAG) I = 3; else I = 2; */INT Top = 1; for (I = 2; I <n; I ++) {While (top & cross (stck [Top-1], stck [Top], p [I]) <= 0) // That's it !!! <= 0 Ah {top --;} stck [++ top] = P [I];} M = Top + 1;} bool CCMP (point a, point B) {if (. y = B. y) return. x <B. x; return. y> B. y;} int main () {int t; scanf ("% d", & T); While (t --) {int CA; int I, J; scanf ("% d", & Ca, & N); for (I = 0; I <n; I ++) scanf ("% d ", & P [I]. x, & P [I]. y); convexhull (); int xx = stck [0]. x, YY = stck [0]. y; int tag = 0; for (I = 1; I <m; I ++) if (stck [I]. y> stck [tag]. Y | (stck [I]. y = stck [tag]. Y & stck [I]. x <stck [tag]. x) Tag = I; printf ("% d \ n", CA, m); for (I = tag; I> = 0; I --) printf ("% d \ n", stck [I]. x, stck [I]. y); for (I = s-1; I> tag; I --) printf ("% d \ n", stck [I]. x, stck [I]. y);} return 0 ;}