由于N不分割0构成的区域，而B分割
所以先搜索有几个由0构成的区域，+1再/2就是答案了
(有没有大佬知道怎么算N的数量)

#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;
}

吐槽一下，这题太坑了，说好的m,n < 30，但我开到50还是炸了