description: You are given N points on an infinite plane with the Cartesian coordinate system on it. N-1 points lay on one line, and one point isn't on that line. You are on point K at the start, and the goal is to visit every point. You can move between any two points in a straight line, and you can revisit points. What is the minimum length of the path? 
  
  Input
  
  The first line contains two integers: N (3 ≤ N ≤ 2*10^5) - the number of points, and K (1 ≤ K ≤ N) - the index of the starting point.
  
  Each of the next N lines contain two integers, A_i, B_i (-10^6 ≤ A_i, B_i ≤ 10^6) - coordinates of the i-th point.
  
  Output
  
  The output contains one number - the shortest path to visit all given points starting from point K. The absolute difference between your solution and the main solution shouldn't exceed 10^-6;
  
  Example
  
  Input
  
  
  5 2
  0 0
  -1 1
  2 -2
  0 1
  -2 2
  
  
  Output
  
  
  7.478709
  
  Note
  
  The shortest path consists of these moves: 
  
  2 -> 5 
  
  5 -> 4 
  
  4 -> 1 
  
  1 -> 3 
  
  There isn't any shorter path possible.
solution: #include <bits/stdc++.h>
  using namespace std;
  int N, M;
  pair<long long, long long> arr[200005], s;
  long double dist(int a, int b) {
    return sqrt((arr[a].first - arr[b].first) * (arr[a].first - arr[b].first) +
                (arr[a].second - arr[b].second) *
                    (arr[a].second - arr[b].second));
  }
  bool chk(int a, int b, int c) {
    return ((arr[b].first - arr[a].first) * (arr[c].second - arr[a].second) -
            (arr[c].first - arr[a].first) * (arr[b].second - arr[a].second)) == 0;
  }
  int main() {
    cin >> N >> M;
    for (int i = 1; i <= N; ++i) {
      cin >> arr[i].first >> arr[i].second;
    }
    s = arr[M];
    if (chk(1, 2, 3)) {
      for (int i = 4; i <= N; ++i) {
        if (!chk(1, 2, i)) {
          swap(arr[i], arr[N]);
          break;
        }
      }
    } else if (N > 3) {
      if (chk(1, 2, 4)) {
        swap(arr[3], arr[N]);
      } else if (chk(1, 3, 4)) {
        swap(arr[2], arr[N]);
      } else {
        swap(arr[1], arr[N]);
      }
    }
    sort(arr + 1, arr + N);
    for (int i = 1; i <= N; ++i) {
      if (arr[i] == s) M = i;
    }
    cout << setprecision(10) << fixed;
    if (M == N) {
      cout << min(dist(1, N), dist(N - 1, N)) + dist(1, N - 1) << endl;
      return 0;
    }
    long double ans = dist(1, N - 1) + min(dist(1, M) + dist(N - 1, N),
                                           dist(M, N - 1) + dist(1, N));
    for (int i = 1; i < N - 1; ++i) {
      int A = i;
      int B = i + 1;
      int C = 1;
      int D = N - 1;
      if (i < M) {
        swap(A, B);
        swap(C, D);
      }
      ans = min(
          ans,
          min(dist(M, A) + dist(A, C) + dist(C, N) + dist(N, B) + dist(B, D),
              dist(M, C) + dist(C, A) + dist(A, N) + dist(N, B) + dist(B, D)));
    }
    cout << ans << endl;
    return 0;
  }
  