【搜索】Flood Fill(洪水填充)算法求网格图中连通块数目，以及求每个连通块内点的个数，DFS与BFS实现

学习资料

1.算法讲解058【必备】洪水填充
2.B10 DFS 水坑计数

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1.求网格图中连通块的数目

P1596 [USACO10OCT] Lake Counting S

八连通坐标写法如下：

DFS写法

#include <iostream>

using namespace std;

const int N = 110;

int n, m, res;
char g[N][N];
int st[N][N];
int dx[8] = {-1, -1, -1, 1, 1, 1, 0, 0};
int dy[8] = {1, 0, -1, 1, 0, -1, 1, -1};

void dfs(int x, int y)
{
    st[x][y] = 1;
    for (int i = 0; i < 8; i++)
    {
        int a = x + dx[i], b = y + dy[i];
        if (a < 0 || a >= n || b < 0 || b >= m) continue;
        if (g[a][b] == '.' || st[a][b]) continue;
        dfs(a, b);
    }
}

int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 0; i < n; i++) cin >> g[i];
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (!st[i][j] && g[i][j] == 'W')
            {
                dfs(i, j);
                res++;
            }
        }
    }
    cout << res;
    return 0;
}

BFS写法

#include <iostream>
#include <queue>

using namespace std;
using PII = pair<int, int>;

const int N = 110;

int n, m, res;
char g[N][N];
int st[N][N];
int dx[8] = {-1, -1, -1, 1, 1, 1, 0, 0};
int dy[8] = {1, 0, -1, 1, 0, -1, 1, -1};

void bfs(int sx, int sy)
{
    queue<PII> q;
    q.push({sx, sy});
    st[sx][sy] = 1;

    while (q.size())
    {
        auto [x, y] = q.front();
        q.pop();

        for (int i = 0; i < 8; i++)
        {
            int a = x + dx[i], b = y + dy[i];
            if (a < 0 || a >= n || b < 0 || b >= m) continue;
            if (st[a][b] || g[a][b] == '.') continue;
            st[a][b] = 1;
            q.push({a, b});
        }
    }
}

int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 0; i < n; i++) cin >> g[i];
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (!st[i][j] && g[i][j] == 'W')
            {
                bfs(i, j);
                res++;
            }
        }
    }
    cout << res;
    return 0;
}

200. 岛屿数量 - 力扣（LeetCode）

DFS写法

class Solution {
public:
    vector<vector<char>> g;
    int n, m, res;
    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
    void dfs(int x, int y)
    {
        g[x][y] = '0';
        for (int i = 0; i < 4; i++)
        {
            int a = x + dx[i], b = y + dy[i];
            if (a < 0 || a >= n || b < 0 || b >= m) continue;
            if (g[a][b] == '0') continue;
            dfs(a, b);
        }
    }
    int numIslands(vector<vector<char>>& grid) {
        g = grid, n = grid.size(), m = grid[0].size();
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                if (g[i][j] == '1')
                {
                    dfs(i, j);
                    res++;
                }
            }
        }
        return res;
    }
};

BFS写法

using PII = pair<int, int>;
class Solution {
public:
    vector<vector<char>> g;
    int res, n, m;
    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
    void bfs(int sx, int sy)
    {
        queue<PII> q;
        q.push({sx, sy});
        g[sx][sy] = '0';

        while (q.size())
        {
            auto [x, y] = q.front();
            q.pop();

            for (int i = 0; i < 4; i++)
            {
                int a = x + dx[i], b = y + dy[i];
                if (a < 0 || a >= n || b < 0 || b >= m) continue;
                if (g[a][b] == '0') continue;
                g[a][b] = '0';
                q.push({a, b});
            }
        }
    }
    int numIslands(vector<vector<char>>& grid) {
        g = grid, n = grid.size(), m = grid[0].size();
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                if (g[i][j] == '1') 
                {
                    bfs(i, j);
                    res++;
                }
            }
        }
        return res;
    }
};

2.求每个连通块内所含点的数目

常见问法有：

（1）求连通块中点的个数的最值

100. 岛屿的最大面积

DFS写法一

#include <iostream>

using namespace std;

const int N = 160;

int n, m, res;
int g[N][N], st[N][N];
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

int dfs(int x, int y)
{
    st[x][y] = 1;
    int cnt = 1; // 当前点也是其所在连通块内的一个点
    for (int i = 0; i < 8; i++)
    {
        int a = x + dx[i], b = y + dy[i];
        if (a < 0 || a >= n || b < 0 || b >= m) continue;
        if (g[a][b] == 0 || st[a][b]) continue;
        cnt += dfs(a, b);
    }
    return cnt;
}

int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++) cin >> g[i][j];

    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (!st[i][j] && g[i][j] == 1)
            {
                res = max(res, dfs(i, j));
            }
        }
    }
    cout << res;
    return 0;
}

DFS写法二

#include <iostream>

using namespace std;

const int N = 160;

int n, m, res, cnt;
int g[N][N], st[N][N];
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

int dfs(int x, int y)
{
    st[x][y] = 1;
    cnt++; // 每次搜时记得在外面把cnt置为0
    for (int i = 0; i < 8; i++)
    {
        int a = x + dx[i], b = y + dy[i];
        if (a < 0 || a >= n || b < 0 || b >= m) continue;
        if (g[a][b] == 0 || st[a][b]) continue;
        dfs(a, b);
    }
    return cnt;
}

int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++) cin >> g[i][j];

    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (!st[i][j] && g[i][j] == 1)
            {
                cnt = 0;//必须要清空cnt，否则计算的是所有连通块内点的个数之和
                res = max(res, dfs(i, j));
            }
        }
    }
    cout << res;
    return 0;
}

DFS写法三

#include <iostream>

using namespace std;

const int N = 160;

int n, m, res, cnt;
int g[N][N], st[N][N];
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

void dfs(int x, int y)
{
    st[x][y] = 1;
    cnt++; // 每次搜时记得在外面把cnt置为0
    for (int i = 0; i < 8; i++)
    {
        int a = x + dx[i], b = y + dy[i];
        if (a < 0 || a >= n || b < 0 || b >= m) continue;
        if (g[a][b] == 0 || st[a][b]) continue;
        dfs(a, b);
    }
}

int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++) cin >> g[i][j];

    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (!st[i][j] && g[i][j] == 1)
            {
                cnt = 0;
                dfs(i, j);
                res = max(res, cnt);
            }
        }
    }
    cout << res;
    return 0;
}

BFS写法

#include <iostream>
#include <queue>

using namespace std;
using PII = pair<int, int>;

const int N = 160;

int n, m, res;
int g[N][N], st[N][N];
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

int bfs(int sx, int sy)
{
    queue<PII> q;
    q.push({sx, sy});
    st[sx][sy] = 1;

    int cnt = 0;
    while (q.size())
    {
        auto [x, y] = q.front();
        q.pop();
        cnt++;

        for (int i = 0; i < 4; i++)
        {
            int a = x + dx[i], b = y + dy[i];
            if (a < 0 || a >= n || b < 0 || b >= m) continue;
            if (g[a][b] == 0 || st[a][b]) continue;
            st[a][b] = 1;
            q.push({a, b});
        }
    }
    return cnt;
}

int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++) cin >> g[i][j];

    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (!st[i][j] && g[i][j] == 1)
            {
                res = max(res, bfs(i, j));
            }
        }
    }
    cout << res;
    return 0;
}

机器人可活动的最大网格点数目

样例1
输入

4 4
1 2 5 2
2 4 4 5
3 5 7 1
4 6 2 4

输出

6

DFS写法

#include <iostream>

using namespace std;

const int N = 160;

int n, m, res;
int g[N][N], st[N][N];
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

int dfs(int x, int y)
{
    st[x][y] = 1;
    int cnt = 1; // 当前点也是其所在连通块内的一个点
    for (int i = 0; i < 8; i++)
    {
        int a = x + dx[i], b = y + dy[i];
        if (a < 0 || a >= n || b < 0 || b >= m) continue;
        if (abs(g[a][b] - g[x][y]) > 1 || st[a][b]) continue;
        cnt += dfs(a, b);
    }
    return cnt;
}

int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++) cin >> g[i][j];

    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (!st[i][j])
            {
                res = max(res, dfs(i, j));
            }
        }
    }
    cout << res;
    return 0;
}

BFS写法

#include <iostream>
#include <queue>

using namespace std;
using PII = pair<int, int>;

const int N = 160;

int n, m, res;
int g[N][N], st[N][N];
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

int bfs(int sx, int sy)
{
    queue<PII> q;
    q.push({sx, sy});
    st[sx][sy] = 1;

    int cnt = 0;//cnt表示坐标为(sx,sy)的点所在连通块包含点的个数
    while (q.size())
    {
        auto [x, y] = q.front();
        q.pop();
        cnt++;

        for (int i = 0; i < 4; i++)
        {
            int a = x + dx[i], b = y + dy[i];
            if (a < 0 || a >= n || b < 0 || b >= m) continue;
            if (abs(g[a][b] - g[x][y]) > 1 || st[a][b]) continue;
            st[a][b] = 1;
            q.push({a, b});
        }
    }
    return cnt;
}

int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++) cin >> g[i][j];

    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (!st[i][j])
            {
                res = max(res, bfs(i, j));
            }
        }
    }
    cout << res;
    return 0;
}

（2）给出某个点坐标，求该点坐标所在连通块包含的点的个数

P1141 01迷宫

如果一个点一个点去搜，数据量较大时会超时，比如本题。
更好的方法是：
对所有连通块编号，用一个一维数组sum[i]存编号为i的连通块所含点的个数。再用一个二维数组id[x][y]存坐标为(x,y)的点所属的连通块的编号。则对于坐标为(x,y)的点，其所在连通块包含的点的个数为sum[id[x][y]]

DFS写法

#include <cstring>
#include <iostream>
#include <queue>

using namespace std;
using PII = pair<int, int>;

const int N = 1010, M = 1e5 + 10;

int n, m, cnt;
char g[N][N];
int st[N][N], id[N][N];
int sum[M]; // 存编号为i的连通块所含点的个数
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

void dfs(int x, int y)
{
    st[x][y] = 1;
    id[x][y] = cnt; // 坐标为(x,y)的点所在连通块的编号为cnt
    sum[cnt]++;     // 编号为cnt的连通块所含点数量+1
    for (int i = 0; i < 4; i++)
    {
        int a = x + dx[i], b = y + dy[i];
        // 本题是n*n矩阵不是n*m矩阵，b>=n不是b>=m，debug几个小时人麻了
        if (a < 0 || a >= n || b < 0 || b >= n) continue;
        if (st[a][b] || g[a][b] == g[x][y]) continue;
        dfs(a, b);
    }
}

int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 0; i < n; i++) cin >> g[i];
    while (m--)
    {
        int x, y;
        cin >> x >> y;
        x--, y--; // 题目下标从1开始，我们存的下标从0开始，所以要-1
        if (!st[x][y])
        {
            dfs(x, y);
            cnt++; // 连通块编号+1
        }
        cout << sum[id[x][y]] << '\n';
    }
    return 0;
}

BFS写法

#include <cstring>
#include <iostream>
#include <queue>

using namespace std;
using PII = pair<int, int>;

const int N = 1010, M = 1e5 + 10;

int n, m, cnt;
char g[N][N];
int st[N][N], id[N][N];
int sum[M]; // 存编号为i的连通块所含点的个数
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

void bfs(int sx, int sy)
{
    queue<PII> q;
    q.push({sx, sy});
    st[sx][sy] = 1;

    id[sx][sy] = cnt; // 坐标为(sx,sy)的点所在的连通块编号为cnt
    while (q.size())
    {
        auto [x, y] = q.front();
        q.pop();
        sum[cnt]++; // 编号为cnt的连通块所含点的数目+1

        for (int i = 0; i < 4; i++)
        {
            int a = x + dx[i], b = y + dy[i];
            if (a < 0 || a >= n || b < 0 || b >= n) continue;
            if (st[a][b] || g[a][b] == g[x][y]) continue;
            id[a][b] = cnt; // 坐标为(a,b)的点所在的连通块编号为cnt
            st[a][b] = 1;
            q.push({a, b});
        }
    }
}

int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 0; i < n; i++) cin >> g[i];
    while (m--)
    {
        int x, y;
        cin >> x >> y;
        x--, y--; // 题目下标从1开始，我们存的下标从0开始，所以要-1
        if (!st[x][y])
        {
            bfs(x, y);
            cnt++; // 连通块编号+1
        }
        cout << sum[id[x][y]] << '\n';
    }
    return 0;
}

---

练习题目

643. 动态网格

687. 扫雷

1097. 池塘计数

本题不是上下左右，而是周围八个格，除去中心

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

typedef pair<int, int> PII;

const int N = 1010;

int n, m;
char g[N][N];
bool st[N][N];
PII q[N * N];

void bfs(int x, int y)
{
    int hh = 0, tt = 0;
    q[0] = {x, y};
    st[x][y] = true;
    
    while (hh <= tt)
    {
        PII t = q[hh ++ ];
        for (int i = t.first - 1; i <= t.first + 1; i ++ )
            for (int j = t.second - 1; j <= t.second + 1; j ++ )
            {
                if (i == t.first && j == t.second) continue;
                if (i < 0 || i >= n || j < 0 || j >= m) continue;
                if (g[i][j] != 'W' || st[i][j]) continue;
                st[i][j] = true;
                q[++ tt] = {i, j};
            }
    }
}

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

1098. 城堡问题

#include <iostream>
#include <algorithm>
#include <cstring>

#define x first
#define y second

using namespace std;

typedef pair<int, int> PII;

const int N = 55, M = N * N;

int n, m;
PII q[M];
int g[N][N];
bool st[N][N];

int bfs(int sx, int sy)
{
    //数组里的x轴正方向是向下的，y轴正方向向右
    int dx[4] = {0, -1, 0, 1}, dy[4] = {-1, 0, 1, 0};

    int hh = 0, tt = 0;
    q[0] = {sx, sy};
    st[sx][sy] = true;

    int area = 0;
    while(hh <= tt)
    {
        PII t = q[hh ++]; //取出队头
        area ++ ;

        for(int i = 0; i < 4; i ++ )
        {
            int a = t.x + dx[i], b = t.y + dy[i];
            if(a < 0 || a >= n || b < 0 || b >= m) continue;
            if(st[a][b]) continue;
            if(g[t.x][t.y] >> i & 1) continue; //如果这个方向是1，说明有墙，不连通

            q[ ++ tt] = {a, b};
            st[a][b] = true;
        }
    }
    return area;
}

int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 0; i < n; i ++ )
        for(int j = 0; j < m; j ++ )
            scanf("%d", &g[i][j]);

    int cnt = 0, area = 0;
    for(int i = 0; i < n; i ++ )
        for(int j = 0; j < m; j ++ )
            if(!st[i][j])
            {
                area = max(area, bfs(i, j));
                cnt ++ ;
            }

    printf("%d\n%d\n", cnt, area);
    return 0;
}

作者：NFYD
链接：https://www.acwing.com/activity/content/code/content/1605169/
来源：AcWing
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1106. 山峰和山谷

#include <iostream>
#include <algorithm>
#include <cstring>

#define x first
#define y second

using namespace std;

typedef pair<int, int> PII;

const int N = 1010, M = N * N;

int n;
PII q[M];
int g[N][N];
bool st[N][N];

void bfs(int sx, int sy, bool &has_higher, bool &has_lower)
{
    int hh = 0, tt = 0;
    q[0] = {sx, sy};
    st[sx][sy] = true;

    while(hh <= tt)
    {
        PII t = q[hh ++];

        //周围八个格都是邻点，按行遍历，特判中心点
        for(int i = t.x - 1; i <= t.x + 1; i ++ )
            for(int j = t.y - 1; j <= t.y + 1; j ++ )
            {
                if(i == t.x && j == t.y) continue; //特判中心点
                if(i < 0 || i >= n || j < 0 || j >= n) continue;
                if(g[i][j] != g[t.x][t.y])
                {
                    if(g[i][j] > g[t.x][t.y]) has_higher = true;
                    else has_lower = true;
                }
                else if(!st[i][j])
                {
                    q[++ tt] = {i, j};
                    st[i][j] = true;
                }
            }
    }
}

int main()
{
    scanf("%d", &n);
    for(int i = 0; i < n; i ++ )
        for(int j = 0; j < n; j ++ )
            scanf("%d", &g[i][j]);

    int peak = 0, valley = 0;
    for(int i = 0; i < n; i ++ )
        for(int j = 0; j < n; j ++ )
            if(!st[i][j])
            {
                bool has_higher = false, has_lower = false;
                bfs(i, j, has_higher, has_lower);
                if(!has_higher) peak ++ ;
                if(!has_lower) valley ++ ;
            }

    printf("%d %d\n", peak, valley);
    return 0;
}

作者：NFYD
链接：https://www.acwing.com/activity/content/code/content/1605426/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

1113. 红与黑

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

typedef pair<int, int> PII;

const int N = 25, M = N * N;

int n, m;
char g[N][N];
PII q[M];
int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};

int bfs(int x, int y)
{
    int hh = 0, tt = 0;
    q[0] = {x, y};
    g[x][y] = '#';
    int res = 0;
    
    while (hh <= tt)
    {
        PII t = q[hh ++ ];
        res ++ ;
        for (int i = 0; i < 4; i ++ )
        {
            int a = t.first + dx[i], b = t.second + dy[i];
            if (a < 0 || a >= n || b < 0 || b >= m) continue;
            if (g[a][b] != '.') continue;
            g[a][b] = '#';
            q[++ tt] = {a, b};
        }
    }
    return res;
}

int main()
{
    while (cin >> m >> n, n || m)
    {
        int x = 0, y = 0;
        for (int i = 0; i < n; i ++ ) cin >> g[i];
        for (int i = 0; i < n; i ++ )
            for (int j = 0; j < m; j ++ )
                if (g[i][j] == '@')
                {
                    x = i, y = j;
                    break;
                }
        printf("%d\n", bfs(x, y));
    }
    return 0;
}

1233. 全球变暖

小变形，需要额外统计其他信息

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

typedef pair<int, int> PII;

const int N = 1010;

int n;
bool st[N][N];
char g[N][N];
PII q[N * N];
int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};

void bfs(int x, int y, int &total, int &bound)
{
    int hh = 0, tt = 0;
    st[x][y] = true;
    q[0] = {x, y};
    
    while (hh <= tt)
    {
        PII t = q[hh ++ ];
        total ++ ;
        bool is_bound = false;
        for (int i = 0; i < 4; i ++ )
        {
            
            int a = t.first + dx[i], b = t.second + dy[i];
            if (a < 0 || a >= n || b < 0 || b >= n || st[a][b]) continue;
            if (g[a][b] == '.')
            {
                is_bound = true;
                continue;
            }
            st[a][b] = true;
            q[++ tt] = {a, b};
        }   
        if (is_bound) bound ++ ;
    }
}

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ ) scanf("%s", g[i]);
    
    int res = 0;
    for (int i = 0; i < n; i ++ )
        for (int j = 0; j < n; j ++ )
            if (g[i][j] == '#' && !st[i][j])
            {
                int total = 0, bound = 0;
                bfs(i, j, total, bound);
                if (total == bound) res ++ ;
            }
    printf("%d", res);
    return 0;
}

1402. 星空之夜

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>

#define x first
#define y second

using namespace std;

typedef pair<int, int> PII;

const int N = 110;

int n, m;
PII q[N * N];
char g[N][N];
int top;

double get_dist(PII a, PII b)
{
    double dx = a.x - b.x;
    double dy = a.y - b.y;
    return sqrt(dx * dx + dy * dy);
}

double get_hash()
{
    double sum = 0;
    for(int i = 0; i < top; i ++ )
        for(int j = i + 1; j < top; j ++ )
            sum += get_dist(q[i], q[j]);
    return sum;
}

char get_id(double key)
{
    static double hash[30];
    static int id = 0;
    for(int i = 0; i < id; i ++ )
        if(fabs(hash[i] - key) < 1e-6)
            return i + 'a';
    hash[id ++ ] = key;
    return id - 1 + 'a';
}

void dfs(int a, int b)
{
    q[top ++ ] = {a, b};
    g[a][b] = '0';
    for(int x = a - 1; x <= a + 1; x ++ )
        for(int y = b - 1; y <= b + 1; y ++ )
        {
            if(x == a && y == b) continue;
            if(x >= 0 && x < n && y >= 0 && y < m && g[x][y] == '1')   
                dfs(x, y);
        }
}

int main()
{
    cin >> m >> n;
    for(int i = 0; i < n; i ++ ) cin >> g[i];
    
    for(int i = 0; i < n; i ++ )
        for(int j = 0; j < m; j ++ )
            if(g[i][j] == '1')
            {
                top = 0;
                dfs(i, j);
                char c = get_id(get_hash());
                for(int k = 0; k < top; k ++ )
                    g[q[k].x][q[k].y] = c;
            }
    
    for(int i = 0; i < n; i ++ ) cout << g[i] << endl;
    return 0;
}

2060. 奶牛选美

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>

#define x first 
#define y second

using namespace std;

typedef pair<int, int> PII;

const int N = 55, M = N * N;

int n, m;
char g[N][N];
vector<PII> points[2];
int dx[] = {-1, 0, 1, 0}, dy[] = {0, -1, 0, 1};
PII q[M]; // 这里M为N * N，因为存的是二维下标，见提高课搜索章节池塘计数
bool st[N][N];

void bfs(int x, int y, vector<PII>& ps)
{
    int hh = 0, tt = 0;
    q[0] = {x, y};
    st[x][y] = true;
    
    while(hh <= tt)
    {
        PII t = q[hh ++];
        ps.push_back(t);
        for(int i = 0; i < 4; i ++ )
        {
            int a = t.x + dx[i], b = t.y + dy[i];
            if(a >= 0 && a < n && b >= 0 && b < m && g[a][b] == 'X' && !st[a][b])
            {
                q[++ tt] = {a, b};
                st[a][b] = true;
            }
        }
    }
}

int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 0; i < n; i ++ ) cin >> g[i];
    
    for(int i = 0, k = 0; i < n; i ++ )
        for(int j = 0; j < m; j ++ )
            if(g[i][j] == 'X' && !st[i][j])
                bfs(i, j, points[k ++ ]);
                
    int res = 1e8;
    for(auto &a : points[0])
        for(auto &b : points[1])
            res = min(res, abs(a.x - b.x) + abs(a.y - b.y) - 1);
    
    cout << res << endl;
    return 0;
}

4004. 传送阵

#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>

#define x first
#define y second

using namespace std;

typedef pair<int, int> PII;

const int N = 55, M = N * N;

int n;
PII q[M];
char g[N][N];
bool st1[N][N], st2[N][N];
int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};

void dfs(int x, int y, bool st[][N])
{
    st[x][y] = true;
    for(int i = 0; i < 4; i ++ )
    {
        int a = x + dx[i], b = y + dy[i];
        if(a >= 0 && a < n && b >= 0 && b < n && !st[a][b] && g[a][b] == '0')
            dfs(a, b, st);
    }
}

int main()
{
    scanf("%d", &n);
    int sx, sy, tx, ty;
    scanf("%d%d%d%d", &sx, &sy, &tx, &ty);
    sx --, sy --, tx --, ty --;
    for(int i = 0; i < n; i ++ ) scanf("%s", g[i]);
    
    dfs(sx, sy, st1);
    if(st1[tx][ty]) puts("0");
    else
    {
        dfs(tx, ty, st2);
        int res = 1e9;
        for(int i = 0; i < n; i ++ )
            for(int j = 0; j < n; j ++ )
                if(st1[i][j])
                    for(int x = 0; x < n; x ++ )
                        for(int y = 0; y < n; y ++ )
                            if(st2[x][y])
                                res = min(res, (i - x) * (i - x) + (j - y) * (j - y));
        printf("%d", res);
    }
    
    return 0;
}