解题思路:
bfs:遍历所有未遍历过的陆地，通过bfs计算出当前位置连通陆地的数量cnt，以及被淹没陆地的数量bound,若cnt == bound表示完整淹没的一个岛屿

dfs:将连通块全部标记，如果这个连通块全部都会淹没，则答案+1，如果这个连通块里面有无法被淹没的，就不+1

bfs代码如下：

#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
typedef pair<int, int>PII;
#define x first
#define y second
const int N = 1010;
char g[N][N];
int res = 0;
bool st[N][N];
int n;int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};void bfs(int x, int y, int &cnt, int &bound) {queue<PII>q;q.push({x, y});st[x][y] = true;while (q.size()) {bool is_bound = false;PII t = q.front();cnt++;q.pop();for (int i = 0; i < 4; i++) {int xx = t.x + dx[i], yy = t.y + dy[i];if (xx < 0 || xx >= n || yy < 0 || yy >= n)continue;if (g[xx][yy] == '.') {is_bound = true;continue;}if (st[xx][yy])continue;st[xx][yy] = true;q.push({xx, yy});}if (is_bound) {bound++;}}}int main() {cin >> n;for (int i = 0; i < n; i++)cin >> g[i];for (int i = 0; i < n; i++)for (int j = 0; j < n; j++) {if (g[i][j] == '#' && !st[i][j]) {int bound = 0, cnt = 0;bfs(i, j, cnt, bound);if (cnt == bound)res++;}}cout << res << endl;return 0;
}

dfs代码如下:

#include <iostream>
using namespace std;
const int N = 1010;
char g[N][N];
int ans;bool flag;int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};
bool vis[N][N];void dfs(int x, int y) {if (g[x + 1][y] == '#' && g[x - 1][y] == '#' && g[x][y + 1] == '#' && g[x][y - 1] == '#')flag = true;///上下左右都是陆地，不会淹没for (int i = 0; i < 4; i++) {int xx = x + dx[i], yy = y + dy[i];if (g[xx][yy] == '#'  && !vis[xx][yy]) {vis[xx][yy] = true;dfs(xx, yy);}}
}int main() {int n;cin >> n;for (int i = 1; i <= n; i++)cin >> g[i];for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++) {if (g[i][j] == '#' && !vis[i][j]) {flag = false;dfs(i, j);if (!flag)ans++;}}cout << ans << endl;return 0;
}