2 条题解

  • 1
    @ 2026-9-7 19:45:43
    using namespace std;
    
    int main() {
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        int n, k;
        cin >> n >> k;
        vector<vector<char>> lit(n, vector<char>(n, 0));
        for (int i = 0; i < k; ++i) {
            int x, y;
            cin >> x >> y;
            --x; --y;
            for (int dx = -1; dx <= 1; ++dx)
                for (int dy = -1; dy <= 1; ++dy) {
                    int nx = x + dx, ny = y + dy;
                    if (0 <= nx && nx < n && 0 <= ny && ny < n)
                        lit[nx][ny] = 1;
                }
        }
        long long cnt = 0;
        for (int i = 0; i < n; ++i)
            cnt += count(lit[i].begin(), lit[i].end(), 1);
        cout << (long long)n * n - cnt << '\n';
        return 0;
    }
    
    
    

    信息

    ID
    3105
    时间
    1000ms
    内存
    256MiB
    难度
    2
    标签
    递交数
    37
    已通过
    15
    上传者