蓝桥杯 全球变暖（BFS）

// #pragma GCC optimize(2)
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <stack>
#include <map>
#include <sstream>
#include <cstring>
#include <set>
#include <cctype>
#include <bitset>
#define IO                       \
	ios::sync_with_stdio(false); \
	// cout.tie(0);
using namespace std;
// int dis[8][2] = {0, 1, 1, 0, 0, -1, -1, 0, 1, -1, 1, 1, -1, 1, -1, -1};
typedef unsigned long long ULL;
typedef long long LL;
typedef pair<int, int> P;
const int maxn = 1e3 + 10;
const int maxm = 2e5 + 10;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const LL mod = 1e9 + 7;
const double pi = acos(-1);
int dis[4][2] = {1, 0, 0, -1, 0, 1, -1, 0};
int m[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
char a[maxn][maxn];
bool vis[maxn][maxn];
int n;
bool judge(int x, int y)
{
	if (x >= 1 && x <= n && y >= 1 && y <= n)
		return true;
	return false;
}
bool check(int x, int y)
{
	if (judge(x - 1, y))
	{
		if (a[x - 1][y] == '.')
			return true;
	}
	if (judge(x + 1, y))
	{
		if (a[x + 1][y] == '.')
			return true;
	}
	if (judge(x, y + 1))
	{
		if (a[x][y + 1] == '.')
			return true;
	}
	if (judge(x, y - 1))
	{
		if (a[x][y - 1] == '.')
			return true;
	}
	return false;
}
bool BFS(int x, int y)
{
	int all = 0;
	int cnt = 0;
	queue<pair<int, int>> q;
	vis[x][y] = true;
	q.push(make_pair(x, y));
	while (!q.empty())
	{
		pair<int, int> p;
		p = q.front();
		q.pop();
		all++;
		int x = p.first;
		int y = p.second;
		if (check(x, y)==true)
			cnt++;
		for (int i = 0; i < 4; i++)
		{
			int tx = x + dis[i][0];
			int ty = y + dis[i][1];
			if (a[tx][ty] == '#' && vis[tx][ty] == false && judge(tx, ty))
			{
				vis[tx][ty] = true;
				q.push(make_pair(tx, ty));
			}
		}
	}
	return cnt == all;
}
int main()
{
#ifdef WXY
	freopen("in.txt", "r", stdin);
	// freopen("out.txt", "w", stdout);
#endif
	IO;
	cin >> n;
	int cnt1 = 0;
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= n; j++)
		{
			cin >> a[i][j];
		}
	}
	check(5, 5);
	int cnt = 0;
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= n; j++)
		{
			if (vis[i][j] == false && a[i][j] == '#')
			{
				if (BFS(i, j))
					cnt++;
			}
		}
	}
	cout << cnt;
	return 0;
}

只需要判断当前连通块（#）中的#会不会被全部淹没，全部淹没，那么岛屿消失，否则不消失。