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AcWing 4959. 岛屿个数（两种解法，通俗解释） - AcWing

1.岛屿个数 - 蓝桥云课 (lanqiao.cn)

代码

#include <bits/stdc++.h>
using namespace std;
#define x first
#define y second
int dx4[4] = {-1, 0, 1, 0}, dy4[4] = {0, 1, 0, -1};
int dx8[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
int dy8[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
using PII = pair<int, int>;
const int N = 55;
char g[N][N];
int cnt, m, n;
void bfs_os(int i, int j) // bfs for outer sea, change 0 to 2
{
    queue<PII> q;
    g[i][j] = 2;
    q.push({i, j});
    while (q.size())
    {
        auto u = q.front();
        q.pop();
        int x = u.x, y = u.y;
        for (int i = 0; i < 8; i++)
        {
            int nx = x + dx8[i];
            int ny = y + dy8[i];
            if (nx < 0 || nx > m + 1 || ny < 0 || ny > n + 1 || g[nx][ny])
                continue;
            g[nx][ny] = 2; //压入之前改标记，提升速度
            q.push({nx, ny});
        }
    }
}
void bfs_is(int i, int j) // bfs for island, change 1 to 0
{
    cnt++;

    queue<PII> q;
    g[i][j] = 0;
    q.push({i, j});
    while (q.size())
    {
        auto u = q.front();
        q.pop();
        int x = u.x, y = u.y;
        for (int i = 0; i < 4; i++)
        {
            int nx = x + dx4[i];
            int ny = y + dy4[i];
            if (nx < 1 || nx > m || ny < 1 || ny > n || g[nx][ny] != 1)
                continue;
            g[nx][ny] = 0;
            q.push({nx, ny});
        }
    }
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    
    int t;
    cin >> t;
    while (t--)
    {
        cnt = 0;
        memset(g, 0, sizeof g);
        cin >> m >> n;
        for (int i = 1; i <= m; i++)
            for (int j = 1; j <= n; j++)
                cin >> g[i][j], g[i][j] -= '0';

        bfs_os(0, 0);
        
        for (int i = 1; i <= m; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                if (g[i][j] == 1 && g[i-1][j] == 2)
                    bfs_is(i, j);
            }
        }
        cout << cnt << '\n';
    }
}