Poj 1113 wall-convex hull

/* The question is not read either. It generally means that a number of points are given, so that a wall is built, including all points, and the distance from all points is not less than l. So the main difference is to find the convex hull. Here are some explanations about the kernel <stdio. h> # include <stdlib. h> # include <math. h> struct point {int X, Y, N;} Dian [1010], STK [1010]; int N; int DIS (point a, point B) {return (. x-b.x) * (. x-b.x) +. y-b.y) * (. y-b.y);} // The returned distance is the square int cross (point P, point S, point E) {return (E. x-s.x) * (p. y-S. Y)-(P. x-s.x) * (E. y-s.y);} int CMP1 (const void * a, const void * B) // sort points in ascending order of values from Y to X, find the {point * c = (point *) A, * D = (point *) B in the lower left corner. If (c-> Y = D-> Y) return C-> X-D-> X; return C-> Y-D-> Y;} int cmp2 (const void * a, const void * B) // sort {int ret; point * c = (point *) A, * D = (point *) B; ret = cross (* D, DIAN [0], * C); If (Ret! = 0) Return-ret; else if (DIS (* C, Dian [0])> DIS (* D, Dian [0]) return 1; else return 0 ;} int tubao (point Dian [], point STK [], int N) // number of input points output by the Input Point (subscript starts from 0) returns the number of vertices on the convex hull {int I, Top = 1; qsort (Dian, N, sizeof (Dian [0]), CMP1); qsort (Dian + 1, n-1, sizeof (Dian [1]), cmp2); STK [0] = Dian [0]; STK [1] = Dian [1]; // STK [2] = Dian [2]; for (I = 2; I <n; ++ I) {While (top> 0 & cross (Dian [I], STK [Top-1], STK [Top]) <= 0) -- top; STK [++ top] = Dian [I];} return top + 1;} I NT main () {int L, I; double ret, A; while (scanf ("% d", & N, & L )! = EOF) {for (I = 0; I <n; I ++) scanf ("% d", & Dian [I]. x, & Dian [I]. y); n = tubao (Dian, STK, n); ret = 3.1415926 * l * 2; for (I = 0; I <n; I ++) {L = DIS (STK [I], STK [(I + 1) % N]); A = SQRT (double) L); RET + = ;} printf ("%. 0f \ n ", RET);} return 0 ;}