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 x[200010], y[200010];
  pair<int, int> p[200010];
  int a[200010];
  long long det(int i, int j, int k) {
    return 1ll * (x[j] - x[i]) * (y[k] - y[i]) -
           1ll * (x[k] - x[i]) * (y[j] - y[i]);
  }
  double dist(int i, int j) {
    return sqrt(1ll * (x[i] - x[j]) * (x[i] - x[j]) +
                1ll * (y[i] - y[j]) * (y[i] - y[j]));
  }
  int mls(int i, int j) { return x[i] == x[j] ? y[i] < y[j] : x[i] < x[j]; }
  int main() {
    int n, k;
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; i++) {
      scanf("%d%d", &x[i], &y[i]);
    }
    --k;
    if (n == 3) {
      double ans = 1e40;
      for (int i = 0; i < n; i++) {
        if (i == k) continue;
        int j = 3 - i - k;
        double t = dist(k, i) + dist(i, j);
        if (ans > t) ans = t;
      }
      printf("%.12f\n", ans);
      return 0;
    }
    int f;
    if (det(0, 1, 2) == 0 || det(0, 1, 3) == 0) {
      for (int i = 2; i < n; i++) {
        if (det(0, 1, i)) f = i;
      }
    } else if (det(0, 2, 1) == 0 || det(0, 2, 3) == 0) {
      for (int i = 1; i < n; i++) {
        if (i != 2 && det(0, 2, i)) f = i;
      }
    } else {
      for (int i = 0; i < n; i++) {
        if (i != 1 && i != 2 && det(1, 2, i)) f = i;
      }
    }
    for (int i = 0; i < n; i++) {
      a[i] = i;
    }
    swap(a[f], a[n - 1]);
    sort(a, a + n - 1, mls);
    double w = dist(a[0], a[n - 2]);
    if (f == k) {
      w += min(dist(a[0], f), dist(a[n - 2], f));
      printf("%.12f\n", w);
      return 0;
    }
    int j;
    for (int i = 0; i < n - 1; i++) {
      if (a[i] == k) j = i;
    }
    double ans = dist(k, f) + min(dist(a[0], f), dist(a[n - 2], f)) + w;
    if (j < n - 2) {
      ans =
          min(ans, w - dist(a[j + 1], a[j]) + dist(a[0], f) + dist(a[n - 2], f));
      ans =
          min(ans, w - dist(a[j + 1], a[j]) + dist(a[0], f) + dist(a[j + 1], f));
    } else {
      ans = min(ans, w + dist(a[0], f));
    }
    if (j) {
      ans =
          min(ans, w - dist(a[j - 1], a[j]) + dist(a[0], f) + dist(a[n - 2], f));
      ans = min(ans,
                w - dist(a[j - 1], a[j]) + dist(a[n - 2], f) + dist(a[j - 1], f));
    } else {
      ans = min(ans, w + dist(a[n - 2], f));
    }
    for (int i = 1; i < n - 2; i++) {
      ans =
          min(ans, w - dist(a[i - 1], a[i]) + dist(f, a[i - 1]) + dist(f, a[i]) +
                       min(dist(a[j], a[0]), dist(a[j], a[n - 2])));
    }
    ans = min(ans, w + min(dist(a[0], f) + dist(a[n - 2], a[j]),
                           dist(a[n - 2], f) + dist(a[0], a[j])));
    printf("%.12f\n", ans);
    return 0;
  }
  