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>
  inline double dis(std::pair<double, double> x, std::pair<double, double> y) {
    return std::sqrt((x.first - y.first) * (x.first - y.first) +
                     (x.second - y.second) * (x.second - y.second));
  }
  inline bool chk(std::pair<double, double> x, std::pair<double, double> y,
                  std::pair<double, double> z) {
    return std::fabs((y.first - x.first) * (z.second - x.second) -
                     (z.first - x.first) * (y.second - x.second)) < 1e-9;
  }
  int n, m;
  std::pair<double, double> a[200005], b;
  int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%lf%lf", &a[i].first, &a[i].second);
    b = a[m];
    if (chk(a[1], a[2], a[3])) {
      for (int i = 4; i <= n; i++)
        if (!chk(a[1], a[2], a[i])) {
          std::swap(a[i], a[n]);
          break;
        }
    } else if (n > 3) {
      if (chk(a[1], a[2], a[4]))
        std::swap(a[3], a[n]);
      else if (chk(a[1], a[3], a[4]))
        std::swap(a[2], a[n]);
      else
        std::swap(a[1], a[n]);
    }
    std::sort(a + 1, a + n);
    for (int i = 1; i <= n; i++)
      if (a[i] == b) m = i;
    if (m == n)
      return printf("%.11lf\n", std::min(dis(a[n], a[1]), dis(a[n], a[n - 1])) +
                                    dis(a[1], a[n - 1])),
             0;
    double ans =
        dis(a[1], a[n - 1]) + std::min(dis(a[m], a[1]) + dis(a[n - 1], a[n]),
                                       dis(a[m], a[n - 1]) + dis(a[1], a[n]));
    for (int i = 1; i < n - 1; i++) {
      std::pair<double, double> A = a[i], B = a[i + 1], C = a[1], D = a[n - 1];
      if (i < m) std::swap(A, B), std::swap(C, D);
      ans = std::min(ans, std::min(dis(a[m], A) + dis(A, C) + dis(C, a[n]) +
                                       dis(a[n], B) + dis(B, D),
                                   dis(a[m], C) + dis(C, A) + dis(A, a[n]) +
                                       dis(a[n], B) + dis(B, D)));
    }
    printf("%.11lf\n", ans);
  }
  