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;
  const int N = 2e5 + 3;
  const double inf = 1e18;
  struct P {
    int x, y;
    int i;
  } p[N];
  double ans = inf;
  using pp = pair<double, double>;
  pp get(int i, int j) {
    if (p[i].x == p[j].x) return {inf, p[i].x};
    double k = (p[i].y - p[j].y) * 1. / (p[i].x - p[j].x),
           b = p[i].y - k * p[i].x;
    return {k, b};
  }
  bool eq(const pp& a, const pp& b) {
    if (fabs(a.first - b.first) < 1e-9 && fabs(a.second - b.second) < 1e-9)
      return 1;
    return 0;
  }
  double dis(int i, int j) { return hypot(p[i].x - p[j].x, p[i].y - p[j].y); }
  int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int i, n, st, x;
    cin >> n >> st;
    for (i = 1; i <= n; ++i) cin >> p[i].x >> p[i].y, p[i].i = i;
    if (n == 3)
      x = 1;
    else {
      if (eq(get(1, 2), get(1, 3))) {
        pp p = get(1, 2);
        for (i = 4; i <= n; ++i)
          if (!eq(get(1, i), p)) {
            x = i;
            goto gg;
          }
        x = 1;
      gg:;
      } else {
        if (eq(get(2, 3), get(3, 4)))
          x = 1;
        else if (eq(get(1, 3), get(3, 4)))
          x = 2;
        else
          x = 3;
      }
    }
    swap(p[x], p[n]);
    sort(p + 1, p + n,
         [](P a, P b) { return a.x < b.x || (a.x == b.x && a.y < b.y); });
    for (i = 1; i <= n; ++i)
      if (p[i].i == st) {
        st = i;
        break;
      }
    if (st == n) {
      ans = min(dis(st, 1), dis(st, n - 1)) + dis(1, n - 1);
    } else {
      ans = min(ans, dis(1, n - 1) + min(dis(st, 1) + dis(n - 1, n),
                                         dis(st, n - 1) + dis(1, n)));
      ans = min(ans, dis(1, n - 1) + dis(st, n) + min(dis(n, n - 1), dis(n, 1)));
      for (i = 2; i <= st; ++i) {
        ans = min(ans, min(dis(st, i) + dis(i, n - 1) + dis(n - 1, n),
                           dis(st, n - 1) + dis(n - 1, i) + dis(i, n)) +
                           dis(n, i - 1) + dis(i - 1, 1));
      }
      for (i = st; i < n - 1; ++i) {
        ans = min(ans, min(dis(st, i) + dis(i, 1) + dis(1, n),
                           dis(st, 1) + dis(1, i) + dis(i, n)) +
                           dis(n, i + 1) + dis(i + 1, n - 1));
      }
    }
    cout << fixed << setprecision(9) << ans;
    exit(0);
    return 0;
  }
  