description: On an endless checkered sheet of paper, n cells are chosen and colored in three colors, where n is divisible by 3. It turns out that there are exactly n/3 marked cells of each of three colors! 
  
  Find the largest such k that it's possible to choose k/3 cells of each color, remove all other marked cells, and then select three rectangles with sides parallel to the grid lines so that the following conditions hold:
  
    * No two rectangles can intersect (but they can share a part of the boundary). In other words, the area of intersection of any two of these rectangles must be 0.
    * The i-th rectangle contains all the chosen cells of the i-th color and no chosen cells of other colors, for i = 1, 2, 3. 
  
  Input
  
  The first line of the input contains a single integer n — the number of the marked cells (3 ≤ n ≤ 10^5, n is divisible by 3).
  
  The i-th of the following n lines contains three integers x_i, y_i, c_i (|x_i|,|y_i| ≤ 10^9; 1 ≤ c_i ≤ 3), where (x_i, y_i) are the coordinates of the i-th marked cell and c_i is its color.
  
  It's guaranteed that all cells (x_i, y_i) in the input are distinct, and that there are exactly n/3 cells of each color.
  
  Output
  
  Output a single integer k — the largest number of cells you can leave.
  
  Examples
  
  Input
  
  
  9
  2 3 1
  4 1 2
  2 1 3
  3 4 1
  5 3 2
  4 4 3
  2 4 1
  5 2 2
  3 5 3
  
  
  Output
  
  
  6
  
  
  Input
  
  
  3
  1 1 1
  2 2 2
  3 3 3
  
  
  Output
  
  
  3
  
  Note
  
  In the first sample, it's possible to leave 6 cells with indexes 1, 5, 6, 7, 8, 9.
  
  In the second sample, it's possible to leave 3 cells with indexes 1, 2, 3.
solution: #include <bits/stdc++.h>
  using namespace std;
  int n, ans, bb, t[10], id[10], l, tot, lim;
  struct item {
    int x, y, c;
  } nx[110000], ny[110000];
  bool cmpx(item a, item b) { return a.x < b.x; }
  bool cmpy(item a, item b) { return a.y < b.y; }
  int scan(int c, int k) {
    tot = 0;
    for (; l <= n; l++) {
      if (nx[l].y <= lim) continue;
      if (nx[l].c == c) {
        tot++;
        if (tot == k) break;
      }
    }
    int f = 0;
    for (int i = l; i <= n; i++) {
      if (nx[i].x != nx[l].x) {
        l = i;
        f = 1;
        break;
      }
    }
    if (f == 0) l = n + 1;
    if (tot != k) return 0;
    return 1;
  }
  int check1(int k) {
    l = 1;
    if (!scan(t[1], k)) return 0;
    if (!scan(t[2], k)) return 0;
    if (!scan(t[3], k)) return 0;
    return 1;
  }
  void binary1(int l, int r) {
    if (l > r) return;
    int mid = l + r >> 1;
    if (check1(mid))
      bb = mid, binary1(mid + 1, r);
    else
      binary1(l, mid - 1);
  }
  int check2(int k) {
    l = 1, tot = 0;
    for (int i = 1; i <= n; i++) {
      if (ny[i].c == t[1]) {
        tot++;
        lim = ny[i].y;
        if (tot == k) break;
      }
    }
    if (!scan(t[2], k)) return 0;
    if (!scan(t[3], k)) return 0;
    return 1;
  }
  void binary2(int l, int r) {
    if (l > r) return;
    int mid = l + r >> 1;
    if (check2(mid))
      bb = mid, binary2(mid + 1, r);
    else
      binary2(l, mid - 1);
  }
  int main() {
    scanf("%d", &n);
    lim = -2e9;
    for (int i = 1; i <= n; i++) scanf("%d%d%d", &nx[i].x, &nx[i].y, &nx[i].c);
    sort(nx + 1, nx + n + 1, cmpx);
    t[1] = 1;
    t[2] = 2;
    t[3] = 3;
    for (int i = 1; i <= 6; i++) {
      binary1(1, n / 3);
      ans = max(ans, bb);
      if (i != 6) next_permutation(t + 1, t + 4);
    }
    t[1] = 1;
    t[2] = 2;
    t[3] = 3;
    for (int i = 1; i <= n; i++) swap(nx[i].x, nx[i].y);
    sort(nx + 1, nx + n + 1, cmpx);
    for (int i = 1; i <= 6; i++) {
      binary1(1, n / 3);
      ans = max(ans, bb);
      if (i != 6) next_permutation(t + 1, t + 4);
    }
    t[1] = 1;
    t[2] = 2;
    t[3] = 3;
    for (int i = 1; i <= n; i++) swap(nx[i].x, nx[i].y);
    memcpy(ny, nx, sizeof(ny));
    sort(nx + 1, nx + n + 1, cmpx);
    sort(ny + 1, ny + n + 1, cmpy);
    for (int i = 1; i <= 6; i++) {
      binary2(1, n / 3);
      ans = max(ans, bb);
      if (i != 6) next_permutation(t + 1, t + 4);
    }
    t[1] = 1;
    t[2] = 2;
    t[3] = 3;
    for (int i = 1; i <= n; i++) nx[i].y = -nx[i].y;
    memcpy(ny, nx, sizeof(ny));
    sort(nx + 1, nx + n + 1, cmpx);
    sort(ny + 1, ny + n + 1, cmpy);
    for (int i = 1; i <= 6; i++) {
      binary2(1, n / 3);
      ans = max(ans, bb);
      if (i != 6) next_permutation(t + 1, t + 4);
    }
    t[1] = 1;
    t[2] = 2;
    t[3] = 3;
    for (int i = 1; i <= n; i++) swap(nx[i].x, nx[i].y);
    memcpy(ny, nx, sizeof(ny));
    sort(nx + 1, nx + n + 1, cmpx);
    sort(ny + 1, ny + n + 1, cmpy);
    for (int i = 1; i <= 6; i++) {
      binary2(1, n / 3);
      ans = max(ans, bb);
      if (i != 6) next_permutation(t + 1, t + 4);
    }
    t[1] = 1;
    t[2] = 2;
    t[3] = 3;
    for (int i = 1; i <= n; i++) nx[i].y = -nx[i].y;
    memcpy(ny, nx, sizeof(ny));
    sort(nx + 1, nx + n + 1, cmpx);
    sort(ny + 1, ny + n + 1, cmpy);
    for (int i = 1; i <= 6; i++) {
      binary2(1, n / 3);
      ans = max(ans, bb);
      if (i != 6) next_permutation(t + 1, t + 4);
    }
    cout << ans * 3;
    return 0;
  }
  