Problem:
Score : 400 points

Problem Statement

We have a grid with H rows and W columns. At first, all cells were painted white.

Snuke painted N of these cells. The i-th ( 1 \leq i \leq N ) cell he painted is the cell at the a_i-th row and b_i-th column.

Compute the following:

For each integer j ( 0 \leq j \leq 9 ), how many subrectangles of size 3×3 of the grid contains exactly j black cells, after Snuke painted N cells?

Constraints

3 \leq H \leq 10^9

3 \leq W \leq 10^9

0 \leq N \leq min(10^5,H×W)

1 \leq a_i \leq H (1 \leq i \leq N)

1 \leq b_i \leq W (1 \leq i \leq N)

(a_i, b_i) \neq (a_j, b_j) (i \neq j)

Input

The input is given from Standard Input in the following format:

H W N
a_1 b_1
:
a_N b_N

Output

Print 10 lines.
The (j+1)-th ( 0 \leq j \leq 9 ) line should contain the number of the subrectangles of size 3×3 of the grid that contains exactly j black cells.

Sample Input 1

4 5 8
1 1
1 4
1 5
2 3
3 1
3 2
3 4
4 4

Sample Output 1

0
0
0
2
4
0
0
0
0
0

There are six subrectangles of size 3×3. Two of them contain three black cells each, and the remaining four contain four black cells each.

Sample Input 2

10 10 20
1 1
1 4
1 9
2 5
3 10
4 2
4 7
5 9
6 4
6 6
6 7
7 1
7 3
7 7
8 1
8 5
8 10
9 2
10 4
10 9

Sample Output 2

4
26
22
10
2
0
0
0
0
0

Sample Input 3

1000000000 1000000000 0

Sample Output 3

999999996000000004
0
0
0
0
0
0
0
0
0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define int long long

#ifdef LOCAL_DEBUG
  #include "LOCAL_DEBUG.hpp"
#endif

signed main(){

  int h,w,n; cin >> h >> w >> n;
  set<pair<int,int>> st,table;
  for(int i = 0; i < n; i++){
    int a,b; cin >> a >> b;
    a--,b--;
    st.insert({a,b});
    for(int dy : {-1,0,1}){
      for(int dx : {-1,0,1}){
        int ny = a + dy;
        int nx = b + dx;
        if(ny-1 < 0 || h <= ny+1 || nx-1 < 0 || w <= nx+1) continue;
        table.insert({ny,nx});
      }
    }
  }

  vector<int> ans(10,0);
  ans[0] = (h-2) * (w-2);
  for(auto p : table){
    int cnt = 0;
    for(int dy : {-1,0,1}){
      for(int dx : {-1,0,1}){
        int ny = p.first + dy;
        int nx = p.second + dx;
        pair<int,int> res = *st.lower_bound(make_pair(ny,nx));
        if(res.first == ny && res.second == nx) cnt++;
        //cnt += st.count(make_pair(ny,nx));
      }
    }
    //cout << p << " " << cnt << endl;
    ans[cnt]++;
    ans[0]--;
  }

  for(int p : ans){
    cout << p << endl;
  }

  return 0;
}