CF-30 D - King's Problem?(枚举+最短路)

D - King's Problem?

Crawling in process... Crawling failed Time Limit:3000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 30D

Description

Every true king during his life must conquer the world, hold the Codeforces world finals, win pink panda in the shooting gallery and travel all over his kingdom.

King Copa has already done the first three things. Now he just needs to travel all over the kingdom. The kingdom is an infinite plane with Cartesian coordinate system on it. Every city is a point on this plane. There are n cities in the kingdom at points with coordinates (x₁, 0), (x₂, 0), ..., (x_n, 0), and there is one city at point (x_n + 1, y_n + 1).

King starts his journey in the city number k. Your task is to find such route for the king, which visits all cities (in any order) and has minimum possible length. It is allowed to visit a city twice. The king can end his journey in any city. Between any pair of cities there is a direct road with length equal to the distance between the corresponding points. No two cities may be located at the same point.

Input

The first line contains two integers n and k (1 ≤ n ≤ 10⁵, 1 ≤ k ≤ n + 1) — amount of cities and index of the starting city. The second line contains n + 1 numbers x_i. The third line contains y_n + 1. All coordinates are integers and do not exceed 10⁶ by absolute value. No two cities coincide.

Output

Output the minimum possible length of the journey. Your answer must have relative or absolute error less than 10^- 6.

Sample Input

Input

3 1
0 1 2 1
1

Output

3.41421356237309490000

Input

3 1
1 0 2 1
1

Output

3.82842712474619030000

Input

4 5
0 5 -1 -5 2
3

Output

14.24264068711928400000

思路：如图，假设将所有点排序后，A为最左点，C为最右点，B为起始点，D为最后的n+1点，E为A到C中的任意一点。

      枚举所有的E点(AE+ED+DE+EC+BC)or(BA+AE+ED+DE+EC)最短路的最优解必在其中。

#include<iostream>
#include<cstring>
#include<iomanip>
#include<algorithm>
#include<cmath>
using namespace std;
const int mm=110000;
double x,y,xx[mm];
int n,m,xxm;
double C(int z)
{
  return hypot(x-xx[z],y);
}
double A(int l,int r)
{
  double z=xx[r]-xx[l]+min(C(l),C(r));
  return z;
}
double B(int l,int r)
{
  double z;
  z=xx[r]-xx[l]+min(C(l)+abs(xxm-xx[r]),C(r)+abs(xxm-xx[l]));
  return z;
}
int main()
{
  while(cin>>n>>m)
  {
    for(int i=0;i<n;i++)
      cin>>xx[i];
    cin>>x>>y;
    xxm=xx[m-1];
    sort(xx,xx+n);
    double ans;
    if(m==n+1)ans=A(0,n-1);
    else 
    {ans=B(0,n-1);
     for(int i=1;i<n;i++)
     {
      double z;z=min(A(0,i-1)+B(i,n-1),A(i,n-1)+B(0,i-1));
      ans=ans<z?ans:z;
      ///cout<<"z="<<z<<endl;
     }
    }
    cout.setf(ios::fixed);
    cout<<setprecision(8)<<ans<<"\n";
  }
}