floodfill

floodfill

1.算法分析

floodfill就是求出连通块的算法,一般可以采用dfs或者bfs,但是dfs容易爆栈,故而采用bfs为佳

2. 例题

acwing1097池塘计数
一块N * M的土地,其中有'W'和'.'组成,'.'为土地,'W'为水,水的八个方向可以连在一起形成池塘,问一块土地上有多少个池塘?
N,M~1e3

#include 

using namespace std;

int const N = 1e3 + 10;
int n, m;
char a[N][N];
int st[N][N];
queue > q;
int dx[] = {-1, -1, 0, 1, 1, 1, 0, -1};
int dy[] = {0, 1, 1, 1, 0, -1, -1, -1};

void bfs(int x, int y) {
    q.push({x, y});
    st[x][y] = 1;
    while (q.size()) {
        auto t = q.front();
        q.pop();
        
        for (int i = 0; i < 8; ++i) {
            int next_x = t.first + dx[i], next_y = t.second + dy[i];
            if (next_x < 0 || next_x > n - 1 || next_y < 0 || next_y > m - 1) continue;
            if (st[next_x][next_y] || a[next_x][next_y] == '.') continue;
            st[next_x][next_y] = 1;
            q.push({next_x, next_y});
        }
    }
}

int main() {
    cin >> n >> m;
    for (int i = 0; i < n; ++i) cin >> a[i];
    
    int res = 0;
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            if (!st[i][j] && a[i][j] == 'W') {
                res ++;
                bfs(i, j);
            }
        }
    }
    cout << res;
    return 0;
}

acwing1106山峰和山谷
N*M的地,每个点上有一个数字,表示高度,如果当前点和周围8个方向的点高度相同,那么可以相连起来成为连通块。如果当前连通块比周围连通块都高,那么该连通块为山峰;如果当前连通块比周围连通块都低,那么该连通块为山谷。求山峰和山谷的数目。
N~1e3,每个点的权值~1e9

#include 

using namespace std;

int const N = 1e3 + 10;
int n, fe, gu;
int a[N][N];
int st[N][N];
queue > q;
int dx[] = {-1, -1, 0, 1, 1, 1, 0, -1};
int dy[] = {0, 1, 1, 1, 0, -1, -1, -1};

void bfs(int x, int y, int val) {
    q.push({x, y});
    st[x][y] = 1;
    while (q.size()) {
        auto t = q.front();
        q.pop();
        
        for (int i = 0; i < 8; ++i) {
            int next_x = t.first + dx[i], next_y = t.second + dy[i];
            if (next_x < 0 || next_x > n - 1 || next_y < 0 || next_y > n - 1) continue;
            if (val > a[next_x][next_y]) fe = 1;
            if (val < a[next_x][next_y]) gu = 1;
            if (st[next_x][next_y] || a[next_x][next_y] != val) continue;
            st[next_x][next_y] = 1;
            q.push({next_x, next_y});
        }
    }
}

int main() {
    cin >> n;
    bool flg = true;  // 判个特例,全部点的值都相同
    int v;
    for (int i = 0; i < n; ++i) 
        for (int j = 0; j < n; ++j) {
            cin >> a[i][j];
            if (!i && !j) v = a[i][j];
            if (a[i][j] != v) flg = false;
        }
    
    if (flg) {
        cout << 1 << " " << 1 << endl;
        return 0;
    }
    
    fe = 0, gu = 0;  // 判断是否出现山峰和山谷
    int cnt_fe = 0, cnt_gu = 0;  // 山峰数目和山谷数目
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            if (!st[i][j]) {
                fe = 0, gu = 0;
                bfs(i, j, a[i][j]);
                if (fe && !gu) cnt_fe++;  // 只有比他高的,那么山峰数加一
                if (!fe && gu) cnt_gu++;  // 只有比他低的,那么山谷数加一
            }
        }
    }
    cout << cnt_fe << " " << cnt_gu;
    return 0;
}

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