题目描述

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样例

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算法1

(暴力枚举) $O(n^2)$

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时间复杂度

参考文献

C++ 代码

#include<iostream>
#include<algorithm>

using namespace std;

const int N = 310;

int n, m, k;
int a[N][N];
int minv[N][N], maxv[N][N];

int main()
{
    int T;
    cin >> T;
    for (int cases = 1; cases <= T; cases ++)
    {
        cin >> n >> m >> k;
        for (int i = 1; i <= n; i ++)
            for (int j = 1; j <= m; j ++)
            {
                cin >> a[i][j];
                minv[i][j] = maxv[i][j] = a[i][j];
            }

        int res = n;
        for (int len = 2; len <= m; len ++)//宽度
            for (int i = 1; i + len - 1 <= m; i ++)//每一行左端点
                for (int j = 1, s = 0; j <= n; j ++)//枚举每一行
                {
                    int &mn = minv[j][i], &mx = maxv[j][i];
                    mn = min(mn, a[j][i + len - 1]);
                    mx = max(mx, a[j][i + len - 1]);

                    if (mx - mn <= k)
                    {
                        s ++;
                        res = max(res, s * len);
                    }
                    else s = 0;
                }

        printf("Case #%d: %d\n", cases, res);
    }
    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

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