1 #include<bits/stdc++.h>
2 using namespace std;
3 const double EPS=1e-6;
4 inline int dcmp(double x){
5 if(fabs(x)<EPS)return 0;
6 else if(x<0)return -1;
7 else return 1;
8 }
9 #define Point Vector
10 struct Vector{
11 double x,y;
12 Vector(double X=0.0,double Y=0.0):x(X),y(Y){}
13 Vector operator+(const Vector &A)const{
14 return Vector(x+A.x,y+A.y);
15 }
16 Vector operator-(const Vector &A)const{
17 return Vector(x-A.x,y-A.y);
18 }
19 double operator*(const Vector &A)const{//点乘·
20 return x*A.x+y*A.y;
21 }
22 Vector operator*(const double &X)const{//数乘
23 return Vector(x*X,y*X);
24 }
25 bool operator==(const Vector&A)const{
26 return ( !dcmp(x-A.x) && !dcmp(y-A.y) );
27 }
28 bool operator<(const Point &A)const{
29 return dcmp(x-A.x) ? x<A.x : y<A.y ;
30 }
31 };
32 inline double polarangle(Vector &A){//向量极角
33 return atan2(A.y,A.x); //?什么函数?
34 }
35 inline double cross(const Vector &A,const Vector &B){//叉乘* （只取数值大小） Arg(A)<Arg(B)时取正值
36 return A.x*B.y-A.y*B.x;
37 }
38 inline double length(const Vector &A){
39 return sqrt(A.x*A.x+A.y*A.y);
40 }
41 inline Point meet(const Point &P1,const Vector &u,const Point &P2,const Vector &v){
42 return P1+u*(cross(P1-P2,v)/cross(v,u));
43 }
44 inline double dis(const Point &A,const Point &P,const Vector &u){
45 return fabs(cross(A-P,u)/length(u));
46 }
47 /*inline double dis(const Point &A,const Point &P,const Point &Q){
48 return fabs(cross(A-P,Q-P)/length(Q-P));
49 }*/
50 const int maxn=1e5+5;
51 Point p[maxn],sta[maxn];
52 int n,k;
53
54 inline void Andrew(){
55 sort(p+1,p+n+1);
56 sta[0]=sta[k=1]=p[1];//放1 or 放2?
57 for(int i=2;i<=n;i++){
58 if(p[i]==p[i-1])continue;//去重
59 while(cross(p[i]-sta[k],sta[k]-sta[k-1])<0)k--;
60 sta[++k]=p[i];
61 }
62 for(int i=n-1;i;i--){//必须循环到1：保证下凸包的正确性
63 if(p[i]==p[i+1])continue;//去重
64 while(cross(p[i]-sta[k],sta[k]-sta[k-1])<0)k--;//凸包边上无点：<=
65 sta[++k]=p[i];
66 }
67 }
68 int main(){
69 scanf("%d",&n);
70 for(int i=1;i<=n;i++){
71 scanf("%lf %lf",&p[i].x,&p[i].y);
72 }
73 Andrew();
74 double ans=0.0;
75 for(int i=1;i<k;i++)ans+=length(sta[i+1]-sta[i]);
76 printf("%.2lf",ans);
77 return 0;
78 }
