题目描述

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样例

blablabla

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

#include<iostream>
#include<queue>

using namespace std;
const int N = 51;
int n, m;
int g[N][N];
bool st[N][N];

int bfs(int x, int y)
{
    int dx[4] = {0, -1, 0, 1}, dy[4] = {-1, 0, 1, 0};
    queue<pair<int, int>> q;
    q.push({x, y});
    st[x][y] = true;
    int res = 0;

    while(q.size())
    {
        auto t = q.front();
        q.pop();
        int x = t.first, y = t.second;
        res ++;
        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 (st[a][b]) continue;
            if(g[x][y] >> i & 1) continue;
            q.push({a, b});
            st[a][b] = true;
        }
    }
    return res;
}

int main()
{
    cin >> n >> m;
    for(int i = 0; i < n; i ++)
        for (int j = 0; j < m; j ++)
            cin >> g[i][j];

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

    cout << cnt << endl << maxv;
    return 0;
}

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla