先贴个题目：

 以及原题链接：1233. 全球变暖 - AcWing题库https://www.acwing.com/problem/content/1235/

 flood fill 算法，和AcWing 1113. 红与黑 解题思路及代码-CSDN博客差不多，但我之前刚开始的思路是先搜索一遍岛屿数量，然后把海平面上升后的地图表示出来然后再搜一遍，二者相减就是最终答案，代码如下：

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;typedef pair<int, int> PII;
#define x first
#define y secondconst int N = 1010;int n;
char map[N][N];
bool sign[N][N];
int dx[4] = {0, 0, -1, 1}, dy[4] = {1, -1, 0, 0};void bfs(int x, int y)
{PII start = make_pair(x, y);sign[x][y] = true;queue<PII> q;q.push(start);while(q.size()){PII tmp = q.front();q.pop();for (int i = 0; i < 4;++i){int tx = tmp.x + dx[i];int ty = tmp.y + dy[i];if(tx<0||tx>=n||ty<0||ty>=n)continue;if(!sign[tx][ty]&&map[tx][ty]=='#'){sign[tx][ty] = true;q.push(make_pair(tx, ty));}}}
}int main()
{cin >> n;for (int i = 0; i < n; ++i)scanf("%s", map[i]);int before = 0, now = 0;for (int i = 0; i < n; ++i) // 搜索原先的岛屿数for (int j = 0; j < n; ++j){if (map[i][j] == '#' && !sign[i][j]){before++;bfs(i, j);}}for (int i = 0; i < n; ++i) // 模拟海平面上升后的岛屿情况for (int j = 0; j < n; ++j){if (map[i][j] == '#'){for (int k = 0; k < 4; ++k){int tx = i + dx[k];int ty = j + dy[k];if (tx < 0 || tx >= n || ty < 0 || ty >= n)continue;if (map[tx][ty] == '.'){map[i][j] = '*';break;}}}}memset(sign, 0, sizeof(sign)); // 重置标记for (int i = 0; i < n; ++i)    // 搜索海平面上升后的岛屿数for (int j = 0; j < n; ++j){if (map[i][j] == '#' && !sign[i][j]){now++;bfs(i, j);}}cout << before - now;return 0;
}

然后发现如果出现：

9
.........
.##.##...
.#####...
.##.##...
.........
.##.#....
.#.###...
.#..#....
.........

这种情况，就会出现错误，错误原因是原来上面那个大岛在被淹没后会变成两个小岛，导致计数出问题，所以不行，于是我就换了个思路，搜索四周没有海域的地块，如果一整块大陆都没有，就让被淹没的岛屿数加一，然后ac了，写了两个版本，bfs和dfs的。

先贴个bfs的：

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;typedef pair<int, int> PII;
#define x first
#define y secondconst int N = 1010;int n;
char map[N][N];
bool sign[N][N];
int dx[4] = {0, 0, -1, 1}, dy[4] = {1, -1, 0, 0};bool bfs(int x, int y)
{bool sign1 = true;PII start = make_pair(x, y);sign[x][y] = true;queue<PII> q;q.push(start);while (q.size()){int cnt = 0;PII tmp = q.front();q.pop();for (int i = 0; i < 4; ++i){int tx = tmp.x + dx[i];int ty = tmp.y + dy[i];if (tx < 0 || tx >= n || ty < 0 || ty >= n)continue;if (!sign[tx][ty] && map[tx][ty] == '#'){sign[tx][ty] = true;q.push(make_pair(tx, ty));}if (map[tx][ty] == '.')cnt++;}if (!cnt)sign1 = false;}return sign1;
}int main()
{cin >> n;for (int i = 0; i < n; ++i)scanf("%s", map[i]);int ans = 0;for (int i = 0; i < n; ++i)for (int j = 0; j < n; ++j){if (map[i][j] == '#' && !sign[i][j]){if (bfs(i, j))ans++;}}cout << ans;return 0;
}

再贴个dfs的：

#include <iostream>
using namespace std;const int N = 1010;int n;
char map[N][N];
bool sign[N][N];
int dx[4] = {0, 0, -1, 1}, dy[4] = {-1, 1, 0, 0};bool dfs(int x, int y)
{bool sign1 = false, sign2 = false;int cnt = 0;for (int i = 0; i < 4; ++i){int tx = x + dx[i];int ty = y + dy[i];if (tx < 0 || tx >= n || ty < 0 || ty >= n)continue;if (map[tx][ty] == '#' && !sign[tx][ty]){sign[tx][ty] = true;sign2 = dfs(tx, ty);sign1 = sign1 || sign2;}if (map[tx][ty] == '.')cnt++;}if (!cnt)sign1 = true;return sign1;
}int main()
{cin >> n;for (int i = 0; i < n; ++i)scanf("%s", map[i]);int ans = 0;for (int i = 0; i < n; ++i)for (int j = 0; j < n; ++j){if (map[i][j] == '#' && !sign[i][j]){sign[i][j] = true;if (!dfs(i, j))ans++;}}cout << ans;return 0;
}

 by————2024.4.7刷题记录