#include <bits/stdc++.h>
using namespace std;
int n, m, g[1005][1005], ans = 0;
void dfs(int x, int y) {
	if (g[x][y] == 0) {
		ans++;
	} else if (g[x][y] == 1 && g[x - 1][y - 1] != 0 && g[x][y - 1] != 0 && g[x + 1][y - 1] != 0 && g[x - 1][y] != 0 && g[x + 1][y] != 0 && g[x - 1][y + 1] != 0 && g[x][y + 1] != 0 && g[x + 1][y + 1] != 0) {
		ans++;
	}
}
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			cin >> g[i][j];
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			if (g[i][j] == 1) {
				g[i][j] = 2;
			} else if (g[i][j] == 0 && g[i - 1][j - 1] == 0 && g[i][j - 1] == 0 && g[i + 1][j - 1] == 0 && g[i - 1][j] == 0 && g[i + 1][j] == 0 && g[i - 1][j + 1] == 0 && g[i][j + 1] == 0 && g[i + 1][j + 1] == 0) {
				g[i][j] = 0;
			} else {
				g[i][j] = 1;
			}
		}
	}
	//2是雷，1是数字，0是空格
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			dfs(i, j);
		}
	}
	cout << ans;
	return 0;
}

我认为思路应该大方向没问题，但是样例输出33