【搜索】Flood Fill 解决连通块问题

算法思路博客： AcWing 1097. 池塘计数

思路：bfs 寻找连通块数量

#include
using namespace std;

typedef long long LL;
const int N = 1e3+10;

int n, m;
char str[N][N];
bool vis[N][N];
int dx[] = {0, 0, 1, 1, 1, -1, -1, -1}, dy[] = {1, -1, 1, 0, -1, 1, 0, -1};

void bfs(int sx, int sy)
{
    queue > que;
    que.push({sx, sy});
    vis[sx][sy] = 1;

    while(!que.empty())
    {
        auto t = que.front(); que.pop();
        int x = it.first, y = it.second;

        for(int i=0; i<8; i++){
            int xx = x + dx[i], yy = y + dy[i];
            if(xx < 0 || xx >= n || yy < 0 || yy >= m || str[xx][yy] == '.' || vis[xx][yy]) continue;
            vis[xx][yy] = 1;
            que.push({xx, yy});
        }
    }
}

int main()
{
    cin >> n >> m;
    for(int i=0; i> str[i];

    int cnt = 0;

    for(int i=0; i