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;
  struct node {
    long long x, y;
    bool operator<(const node a) const {
      if (x == a.x) return y < a.y;
      return x < a.x;
    }
    bool operator==(const node a) const { return x == a.x && y == a.y; }
  } mp[200005], ts;
  long double dis(int i, int j) {
    node a = mp[i], b = mp[j];
    long long sx = a.x - b.x, sy = a.y - b.y;
    return sqrt(sx * sx + sy * sy);
  }
  long double dis(vector<int> v) {
    int las = 0;
    long double s = 0;
    for (auto it = v.begin(); it != v.end(); it++) {
      if (las) s += dis(las, *it);
      las = *it;
    }
    return s;
  }
  int n, st;
  bool chk(node a, node b, node s) {
    return (a.y - s.y) * (b.x - s.x) == (b.y - s.y) * (a.x - s.x);
  }
  int main() {
    scanf("%d%d", &n, &st);
    for (int i = 1; i <= n; i++) scanf("%lld%lld", &mp[i].x, &mp[i].y);
    if (n == 3) {
      int xx = 1 + (st == 1), yy = 2 + (st <= 2);
      printf("%.9Lf", min(dis(st, xx) + dis(xx, yy), dis(st, yy) + dis(yy, xx)));
    }
    ts = mp[st];
    sort(mp + 1, mp + n + 1);
    n = unique(mp + 1, mp + n + 1) - mp - 1;
    if (!chk(mp[2], mp[3], mp[1]) && !chk(mp[2], mp[4], mp[1]) &&
        !chk(mp[3], mp[4], mp[1]))
      swap(mp[1], mp[n]);
    else if (!chk(mp[1], mp[3], mp[2]) && !chk(mp[1], mp[4], mp[2]) &&
             !chk(mp[3], mp[4], mp[2]))
      swap(mp[2], mp[n]);
    else
      for (int i = 3; i <= n; i++)
        if (!chk(mp[1], mp[2], mp[i])) swap(mp[i], mp[n]);
    sort(mp + 1, mp + n);
    st = find(mp + 1, mp + n + 1, ts) - mp;
    if (st == n)
      printf("%.9Lf", dis(1, n - 1) + min(dis(1, n), dis(n - 1, n)));
    else {
      long double ans = 1e18;
      for (int i = 1; i <= n - 2; i++)
        ans = min(
            {dis({st, i, 1, n, i + 1, n - 1}), dis({st, i + 1, n - 1, n, i, 1}),
             dis({st, i, n}) + dis(1, n - 1) + min(dis(n, 1), dis(n, n - 1)),
             dis({1, i, n, i + 1, n - 1}) + min(dis(st, 1), dis(st, n - 1)),
             ans});
      printf("%.9Lf",
             min(min(dis(st, 1) + dis(n - 1, n), dis(st, n - 1) + dis(1, n)) +
                     dis(1, n - 1),
                 ans));
    }
    return 0;
  }
  