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 pt {
    int x, y, id;
    pt() {}
    pt(int x, int y) : x(x), y(y) {}
  } a[1000005];
  bool operator<(const pt a, const pt b) {
    return a.x == b.x ? a.y < b.y : a.x < b.x;
  }
  pt operator-(const pt a, const pt b) { return pt(a.x - b.x, a.y - b.y); }
  long long Cross(const pt a, const pt b) {
    return 1ll * a.x * b.y - 1ll * a.y * b.x;
  }
  double F(int i, int j) {
    return sqrt(1ll * (a[i].x - a[j].x) * (a[i].x - a[j].x) +
                1ll * (a[i].y - a[j].y) * (a[i].y - a[j].y));
  }
  int n, k;
  int main() {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) scanf("%d%d", &a[i].x, &a[i].y), a[i].id = i;
    if (n > 3) {
      long long ans1 = Cross(a[2] - a[1], a[3] - a[1]),
                ans2 = Cross(a[4] - a[1], a[3] - a[1]);
      long long ans3 = Cross(a[2] - a[1], a[4] - a[1]);
      if (ans1 && ans2 && ans3)
        swap(a[1], a[n]);
      else if (ans1 && ans3)
        swap(a[2], a[n]);
      else if (ans1 && ans2)
        swap(a[3], a[n]);
      else if (ans2 && ans3)
        swap(a[4], a[n]);
      else {
        for (int i = 3; i <= n; i++)
          if (Cross(a[2] - a[1], a[i] - a[1])) {
            swap(a[i], a[n]);
            break;
          }
      }
    }
    sort(a + 1, a + n);
    for (int i = 1; i <= n; i++)
      if (a[i].id == k) {
        k = i;
        break;
      }
    double ans = 1e18;
    if (k == n) {
      ans = min(ans, F(1, n - 1) + min(F(1, n), F(n - 1, n)));
      printf("%.10f\n", ans);
      return 0;
    }
    ans = min(ans, F(k, n) + F(1, n - 1) + min(F(1, n), F(n - 1, n)));
    ans =
        min(ans, F(1, n - 1) + min(F(1, n) + F(n - 1, k), F(n - 1, n) + F(1, k)));
    ans = min(ans, F(k, 1) + F(1, n) +
                       (k == n - 1 ? 0 : F(n, k + 1) + F(k + 1, n - 1)));
    ans = min(ans, F(k, n - 1) + F(n - 1, n) +
                       (k == 1 ? 0 : F(n, k - 1) + F(k - 1, 1)));
    for (int i = 2; i <= k; i++)
      ans = min(ans, F(1, i - 1) + min(F(i - 1, n), F(1, n)) + F(i, n) +
                         F(i, n - 1) + F(k, n - 1));
    for (int i = k; i < n - 1; i++)
      ans = min(ans, F(k, 1) + F(1, i) + F(i, n) + min(F(n, i + 1), F(n, n - 1)) +
                         F(i + 1, n - 1));
    printf("%.10f\n", ans);
    return 0;
  }
  