here

主要是看边界的情况，因为他说最边上的距离不能超过L这就意味着拐角处外面的城墙应该是属于弧状的。
这样最后的结果就会是凸包周长加上 2 Π L 2ΠL 2ΠL
套一下板子就好了。

ac code

#include <bits/stdc++.h>
#include <iostream>
#include<algorithm>
#include <cstdio>
#include <cmath>
#define re register int
#define int long long
#define inf 0x3f
using namespace std;
const int N=1e5+5;
int n,top,ans;
struct Point{
	double x,y;
}point[N],stk[N];

double dis(Point a,Point b)
{
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

double cross(Point a,Point b,Point c){
  return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);
}

void find_miny(){
   Point tmp=point[0];
   int flag=0;
   for(int i=1;i<n;i++){
     if(point[i].y<tmp.y||(point[i].y==tmp.y&&point[i].x<tmp.x)){
        tmp=point[i];
        flag=i;
     }
   }
   if(flag) swap(point[0],point[flag]);
}

bool cmp(Point a,Point b){
  double n=cross(point[0],a,b);
  if(n>0||(n==0&&dis(point[0],a)>dis(point[0],b))) return true;
  return false;
}

void Graham(){
  top=-1;
  stk[++top]=point[0]; stk[++top]=point[1];
  for(int i=2;i<n;i++){
    while(top&&cross(stk[top-1],stk[top],point[i])<0) top--;
    stk[++top]=point[i];
  }
}
void solve()
{
    double l;
	cin>>n>>l;
	top=0;
	memset(stk,0,sizeof(stk));
	memset(point,0,sizeof(point));
	for(int i=0;i<n;i++){
		cin>>point[i].x>>point[i].y;
    }

    find_miny();
    sort(point+1,point+n,cmp);
    Graham();
    stk[++top]=point[0];

//    for(int i=0;i<=top;i++)
//    {
//    	cout<<stk[i].x<<" "<<stk[i].y<<endl;
//	}

    double fans=2.0*l*acos(-1);
    for(int i=0;i<top;i++)
    {
    	fans+=dis(stk[i],stk[i+1]);
	}
	if(n==1)
	{
		cout<<0<<endl;
		return;
	}
	if(top==2)
	{
		printf("%.0f\n",fans/2);
		return;
	}
    printf("%.0f\n",fans);
}
signed main() {
	solve();
	return 0;
}