这道题和老师讲过的没什么两样， 被洪水淹没的地区肯定和边缘相连， 所以从边缘出发做bfs就行了。

#include <bits/stdc++.h>
using namespace std;
struct Point{
    int x, y;
};
queue<Point> q;
int n, m, dx[4]={1, 0, 0, -1}, dy[4]={0, 1, -1, 0}, diff[505][505], sum;
char a[505][505];
void bfs(){
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            if ((i==1 || i==n || j==1 || j==m) && a[i][j] == '0'){
                q.push((Point){i, j});
                diff[i][j] = 1;
            }
    while (q.size()){
        Point now = q.front();
        q.pop();
        for (int i = 0; i < 4; i++){
            int nx = now.x + dx[i], ny = now.y + dy[i];
            if (nx>=1&&nx<=n&&ny>=1&&ny<=m&&a[nx][ny]=='0'&&!diff[nx][ny]){
                diff[nx][ny] = 1;
                q.push((Point){nx, ny});
            }
        }
    }
}
int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            cin >> a[i][j];
    bfs();
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            if (a[i][j] == '0' && diff[i][j] == 0)
                sum++;
    cout << sum;
    return 0;
}