Poj 1113 wall convex hull Application

Link: poj 1113 monotonic link convex bag Summary

Question: we still use a convex bag to find the shortest perimeter. We only need to add the length of a circle. We can apply the template to solve this problem;

AC code:

1 # include <iostream> 2 # include <algorithm> 3 # include <cstdio> 4 # include <cmath> 5 using namespace STD; 6 int M, N; 7 struct p 8 {9 Double X, Y; 10 friend int operator <(p a, p B) 11 {12 if (. x <B. X | (. X = B. X &. Y <B. y) 13 return 1; 14 return 0; 15} 16} A [1010], ANS [1010]; 17 double judge (p a, p B, P C) 18 {19 Return (B. x-a.x) * (C. y-a.y)-(C. x-a.x) * (B. y-a.y); 20} 21 int getnumber () // find the key point to find the shortest perimeter 22 {23 int K = 0; 24 for (INT I = 0; I <m; I ++) 25 {26 while (k> 1 & judge (ANS [K-2], ANS [k-1], a [I]) <= 0) 27 {28 K --; 29} 30 ans [k ++] = A [I]; 31} 32 int k1 = K; 33 for (INT I = s-1; i> = 0; I --) 34 {35 while (k> K1 & judge (ANS [K-2], ANS [k-1], a [I]) <= 0) 36 {37 k --; 38} 39 ans [K + +] = A [I]; 40} 41 return K-1; // The first vertex is saved twice; 42} 43 44 int main () 45 {46 scanf ("% d", & M, & N); 47 for (INT I = 0; I <m; I ++) 48 scanf ("% lf", & A [I]. x, & A [I]. y); 49 sort (A, A + M); 50 int Top = getnumber (); 51 double ans1 = 0; 52 int I; 53 for (I = 0; I <top; I ++) {54 ans1 + = SQRT (ANS [I]. x-ans [I + 1]. x) * (ANS [I]. x-ans [I + 1]. x) + (ANS [I]. y-ans [I + 1]. y) * (ANS [I]. y-ans [I + 1]. y); 55} 56 // In fact, you only need to add a circle perimeter to the end, and the outer angle of the polygon is 360 degrees; 57 ans1 + = 2*3.141592653 * N; 58 printf ("%. 0f \ n ", ans1); // G ++ (. 0l %), c ++ can return 0; 60}

 

 

Poj 1113 wall convex hull Application