#include<bits/stdc++.h>
#define N 1010
using namespace std;
char b[N][N];
bool vis[N][N];
int p[8][3] = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}, {1, 1}, {-1, -1}, {-1, 1}, {1, -1}}, m, n;
struct node{
    int x, y;
}q[N];
void bfs(int x, int y){
    int head = 1, tail = 1;
    q[tail].x = x; q[tail].y = y;
    tail++;
    while(head < tail){
        vis[q[head].x][q[head].y] = 1;
        for(int i = 0; i < 8; i++){
            int nx = q[head].x + p[i][0], ny = q[head].y + p[i][1];
            if(nx >= 0 && nx <= m+1  &&  ny >= 0 && ny <= n+1  &&  b[nx][ny] == '0'  &&  !vis[nx][ny]){
                q[tail].x = nx; q[tail].y = ny;  //入队
                tail++;
                vis[nx][ny] = 1;
            }
        }
        head++;  //队头出队 
    }
}
int main(){
    ios::sync_with_stdio(false);
    int cnt = 0;
    cin >> m >> n;
    memset(vis, 0, sizeof(vis)); memset(b, 0, sizeof(b));
    for(int i = 1; i <= m; i++){
        for(int j = 1; j <= n; j++)cin >> b[i][j];
    }
    for(int i = 1; i <= m; i++){
        for(int j = 1; j <= n; j++){
            if(b[i][j] == '0' && !vis[i][j]){
                cnt++;
                bfs(i, j);
            }
        }
    }
    cout << (cnt-1)/2;
    return 0;
}

/*

*/
#define method_1
#ifdef method_1
/*
100分算法 
种子填充 找到有几个含0的联通块 记录其个数为cnt 
则答案就是cnt-1后除以二 
*/
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long ll;
const int maxn=100+5;
const ll INF=0x3f3f3f3f3f3f3f3fll;
const int dx[]= {0,-1,0,1};
const int dy[]= {1,0,-1,0};
int n,m;
char a[maxn][maxn];
int vis[maxn][maxn];
bool check(int x,int y) {
    if(x<0||x>(n+1)||y<0||y>(m+1)) return false;
    return true;
}
void dfs(int x,int y,int c) {
    if(check(x,y)==false) return;
    if(vis[x][y]||(a[x][y]!='0')) return;
    vis[x][y]=c;
    for(int i=0; i<=3; i++) {
        int newx=x+dx[i];
        int newy=y+dy[i];
        //  if(check(newx,newy)&&(vis[newx][newy]==0)&&a[newx][newy]=='0'){
        //  vis[newx][newy]=c;
        dfs(newx,newy,c);
        //  }
    }
}
int main() {
    ios::sync_with_stdio(false);
    freopen("偷拍硕哥3.in","r",stdin);
    for(int i=0; i<=n+1; i++) {
        for(int j=0; j<=m+1; j++) {
            a[i][j]='0';
            vis[i][j]=0;
        }
    }
    cin>>n>>m;
    for(int i=1; i<=n; i++) {
        for(int j=1; j<=m; j++) {
            cin>>a[i][j];
        }
    }
    /*
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            cout<<a[i][j];
        }
        cout<<endl;
    }
    */
    int cnt=0;
    for(int i=0; i<=n+1; i++) {
        for(int j=0; j<=m+1; j++) {
            if((a[i][j]=='0')&&(vis[i][j]==0)) {
                dfs(i,j,++cnt);
            }
        }
    }
    /* 
    for(int i=0; i<=n+1; i++) {
        for(int j=0; j<=m+1; j++) {
            cout<<vis[i][j];
        }
        cout<<endl;
    }
    */ 
    cout<<(cnt-1)/2;
    return 0;
}
#endif
#ifdef method_2
/*

*/

#endif
#ifdef method_3
/*

*/

#endif

/*

/
#define method_1
#ifdef method_1
/
100分算法
种子填充 找到有几个含0的联通块 记录其个数为cnt
则答案就是cnt-1后除以二
/
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long ll;
const int maxn=100+5;
const ll INF=0x3f3f3f3f3f3f3f3fll;
const int dx[]= {0,-1,0,1};
const int dy[]= {1,0,-1,0};
int n,m;
char a[maxn][maxn];
int vis[maxn][maxn];
bool check(int x,int y) {
if(x<0||x>(n+1)||y<0||y>(m+1)) return false;
return true;
}
void dfs(int x,int y,int c) {
if(check(x,y)==false) return;
if(vis[x][y]||(a[x][y]!='0')) return;
vis[x][y]=c;
for(int i=0; i<=3; i++) {
int newx=x+dx[i];
int newy=y+dy[i];
// if(check(newx,newy)&&(vis[newx][newy]==0)&&a[newx][newy]=='0'){
// vis[newx][newy]=c;
dfs(newx,newy,c);
// }
}
}
int main() {
ios::sync_with_stdio(false);
freopen("偷拍硕哥3.in","r",stdin);
for(int i=0; i<=n+1; i++) {
for(int j=0; j<=m+1; j++) {
a[i][j]='0';
vis[i][j]=0;
}
}
cin>>n>>m;
for(int i=1; i<=n; i++) {
for(int j=1; j<=m; j++) {
cin>>a[i][j];
}
}
/
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
cout<<a[i][j];
}
cout<<endl;
}
/
int cnt=0;
for(int i=0; i<=n+1; i++) {
for(int j=0; j<=m+1; j++) {
if((a[i][j]=='0')&&(vis[i][j]==0)) {
dfs(i,j,++cnt);
}
}
}
/
for(int i=0; i<=n+1; i++) {
for(int j=0; j<=m+1; j++) {
cout<<vis[i][j];
}
cout<<endl;
}
/
cout<<(cnt-1)/2;
return 0;
}
#endif
#ifdef method_2
/

*/

#endif
#ifdef method_3
/*

*/

#endif