Problem:
Score : 700 points

Problem Statement

Tonight, in your favourite cinema they are giving the movie Joker and all seats are occupied. In the cinema there are N rows with N seats each, forming an N\times N square. We denote with 1, 2,\dots, N the viewers in the first row (from left to right); with N+1, \dots, 2N the viewers in the second row (from left to right); and so on until the last row, whose viewers are denoted by N^2-N+1,\dots, N^2.

At the end of the movie, the viewers go out of the cinema in a certain order: the i-th viewer leaving her seat is the one denoted by the number P_i. The viewer P_{i+1} waits until viewer P_i has left the cinema before leaving her seat.
To exit from the cinema, a viewer must move from seat to seat until she exits the square of seats (any side of the square is a valid exit). A viewer can move from a seat to one of its 4 adjacent seats (same row or same column).
While leaving the cinema, it might be that a certain viewer x goes through a seat currently occupied by viewer y; in that case viewer y will hate viewer x forever. Each viewer chooses the way that minimizes the number of viewers that will hate her forever.

Compute the number of pairs of viewers (x, y) such that y will hate x forever.

Constraints

2 \le N \le 500

The sequence P_1, P_2, \dots, P_{N^2} is a permutation of \{1, 2, \dots, N^2\}.

Input

The input is given from Standard Input in the format

N
P_1 P_2 \cdots P_{N^2}

Output

If ans is the number of pairs of viewers described in the statement, you should print on Standard Output

ans

Sample Input 1

3
1 3 7 9 5 4 8 6 2

Sample Output 1

1

Before the end of the movie, the viewers are arranged in the cinema as follows:

1 2 3
4 5 6
7 8 9

The first four viewers leaving the cinema (1, 3, 7, 9) can leave the cinema without going through any seat, so they will not be hated by anybody.

Then, viewer 5 must go through one of the seats where viewers 2, 4, 6, 8 are currently seated while leaving the cinema; hence he will be hated by at least one of those viewers.

Finally the remaining viewers can leave the cinema (in the order 4, 8, 6, 2) without going through any occupied seat (actually, they can leave the cinema without going through any seat at all).

Sample Input 2

4
6 7 1 4 13 16 10 9 5 11 12 14 15 2 3 8

Sample Output 2

3

Sample Input 3

6
11 21 35 22 7 36 27 34 8 20 15 13 16 1 24 3 2 17 26 9 18 32 31 23 19 14 4 25 10 29 28 33 12 6 5 30

Sample Output 3

11

Solution in Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)


import java.util.Arrays;

public class Main {

  private static void solve() {
    int n = ni();
    int[] p = na(n * n);

    int[][] map = new int[n][n];
    int ptr = 0;
    for (int i = 0; i < n * n; i++) {
      p[i]--;
      int y = p[i] / n;
      int x = p[i] % n;

      map[y][x] = ptr++;
    }

    long ret = 0;
    final int[] td = new int[n * n];
    int[] from = new int[2];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        from[0] = i;
        from[1] = j;
        int now = dijk(map, from, td);
        ret += now;
      }
    }
    System.out.println(ret);
  }

  static int[][] dir = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

  public static int dijk(int[][] map, int[] from, final int[] td) {
    int n = map.length;

    Arrays.fill(td, 999999999);
    MinHeap q = new MinHeap(n * n);


    td[from[0] * n + from[1]] = 0;
    q.add(from[0] * n + from[1], 0);
    int v = map[from[0]][from[1]];

    while (q.size() > 0) {
      int cur = q.argmin();
      q.remove(cur);

      int cy = cur / n;
      int cx = cur % n;

      for (int[] d : dir) {
        int nx = cx + d[0];
        int ny = cy + d[1];
        if (nx < 0 || nx >= n || ny < 0 || ny >= n)
          return td[cur];

        int nd = td[cur] + (v < map[ny][nx] ? 1 : 0);
        int next = nx + ny * n;
        if (nd < td[next]) {
          q.remove(next);
          td[next] = nd;
          q.update(next, nd);
        }
      }
    }
    return -1;
  }



  public static class MinHeap {
    public int[] a;
    public int[] map;
    public int[] imap;
    public int n;
    public int pos;
    public static int INF = Integer.MAX_VALUE;

    public MinHeap(int m) {
      n = m + 2;
      a = new int[n];
      map = new int[n];
      imap = new int[n];
      Arrays.fill(a, INF);
      Arrays.fill(map, -1);
      Arrays.fill(imap, -1);
      pos = 1;
    }

    public int add(int ind, int x) {
      int ret = imap[ind];
      if (imap[ind] < 0) {
        a[pos] = x;
        map[pos] = ind;
        imap[ind] = pos;
        pos++;
        up(pos - 1);
      }
      return ret != -1 ? a[ret] : x;
    }

    public int update(int ind, int x) {
      int ret = imap[ind];
      if (imap[ind] < 0) {
        a[pos] = x;
        map[pos] = ind;
        imap[ind] = pos;
        pos++;
        up(pos - 1);
      } else {
        int o = a[ret];
        a[ret] = x;
        up(ret);
        down(ret);
        // if(a[ret] > o){
        // up(ret);
        // }else{
        // down(ret);
        // }
      }
      return x;
    }

    public int remove(int ind) {
      if (pos == 1)
        return INF;
      if (imap[ind] == -1)
        return INF;

      pos--;
      int rem = imap[ind];
      int ret = a[rem];
      map[rem] = map[pos];
      imap[map[pos]] = rem;
      imap[ind] = -1;
      a[rem] = a[pos];
      a[pos] = INF;
      map[pos] = -1;

      up(rem);
      down(rem);
      return ret;
    }

    public int min() {
      return a[1];
    }

    public int argmin() {
      return map[1];
    }

    public int size() {
      return pos - 1;
    }

    private void up(int cur) {
      for (int c = cur, p = c >>> 1; p >= 1 && a[p] > a[c]; c >>>= 1, p >>>= 1) {
        int d = a[p];
        a[p] = a[c];
        a[c] = d;
        int e = imap[map[p]];
        imap[map[p]] = imap[map[c]];
        imap[map[c]] = e;
        e = map[p];
        map[p] = map[c];
        map[c] = e;
      }
    }

    private void down(int cur) {
      for (int c = cur; 2 * c < pos;) {
        int b = a[2 * c] < a[2 * c + 1] ? 2 * c : 2 * c + 1;
        if (a[b] < a[c]) {
          int d = a[c];
          a[c] = a[b];
          a[b] = d;
          int e = imap[map[c]];
          imap[map[c]] = imap[map[b]];
          imap[map[b]] = e;
          e = map[c];
          map[c] = map[b];
          map[b] = e;
          c = b;
        } else {
          break;
        }
      }
    }
  }


  public static void main(String[] args) {
    new Thread(null, new Runnable() {
      @Override
      public void run() {
        long start = System.currentTimeMillis();
        String debug = args.length > 0 ? args[0] : null;
        if (debug != null) {
          try {
            is = java.nio.file.Files.newInputStream(java.nio.file.Paths.get(debug));
          } catch (Exception e) {
            throw new RuntimeException(e);
          }
        }
        reader = new java.io.BufferedReader(new java.io.InputStreamReader(is), 32768);
        solve();
        out.flush();
        tr((System.currentTimeMillis() - start) + "ms");
      }
    }, "", 64000000).start();
  }

  private static java.io.InputStream is = System.in;
  private static java.io.PrintWriter out = new java.io.PrintWriter(System.out);
  private static java.util.StringTokenizer tokenizer = null;
  private static java.io.BufferedReader reader;

  public static String next() {
    while (tokenizer == null || !tokenizer.hasMoreTokens()) {
      try {
        tokenizer = new java.util.StringTokenizer(reader.readLine());
      } catch (Exception e) {
        throw new RuntimeException(e);
      }
    }
    return tokenizer.nextToken();
  }

  private static double nd() {
    return Double.parseDouble(next());
  }

  private static long nl() {
    return Long.parseLong(next());
  }

  private static int[] na(int n) {
    int[] a = new int[n];
    for (int i = 0; i < n; i++)
      a[i] = ni();
    return a;
  }

  private static char[] ns() {
    return next().toCharArray();
  }

  private static long[] nal(int n) {
    long[] a = new long[n];
    for (int i = 0; i < n; i++)
      a[i] = nl();
    return a;
  }

  private static int[][] ntable(int n, int m) {
    int[][] table = new int[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        table[i][j] = ni();
      }
    }
    return table;
  }

  private static int[][] nlist(int n, int m) {
    int[][] table = new int[m][n];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        table[j][i] = ni();
      }
    }
    return table;
  }

  private static int ni() {
    return Integer.parseInt(next());
  }

  private static void tr(Object... o) {
    if (is != System.in)
      System.out.println(java.util.Arrays.deepToString(o));
  }
}