和玉蟾宮很像，條件改成不相等就行了。

懸線法題目 洛谷 P1169  p4147  p2701  p1387 

#include<cstdio>
#include<algorithm>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
#define _for(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;

const int MAXN = 2123;
int up[MAXN][MAXN], l[MAXN][MAXN];
int r[MAXN][MAXN], a[MAXN][MAXN], n, m;

int main()
{
	scanf("%d%d", &n, &m);
	_for(i, 1, n)
		_for(j, 1, m)
		{
			scanf("%d", &a[i][j]);
			l[i][j] = r[i][j] = j;
			up[i][j] = 1;
		} 
	
	_for(i, 1, n)
	{
		_for(j, 2, m)
			if(a[i][j] != a[i][j-1])
				l[i][j] = l[i][j-1];
		for(int j = m - 1; j >= 1; j--)
			if(a[i][j] != a[i][j+1])
				r[i][j] = r[i][j+1];
	}
	
	int ans1 = 0, ans2 = 0;
	_for(i, 1, n)
		_for(j, 1, m)
		{
			if(i > 1 && a[i][j] != a[i-1][j])
			{
				up[i][j] = up[i-1][j] + 1;
				l[i][j] = max(l[i][j], l[i-1][j]);
				r[i][j] = min(r[i][j], r[i-1][j]);
			}
			int w = r[i][j] - l[i][j] + 1, h = up[i][j];
			ans1 = max(ans1, min(w, h) * min(w, h));
			ans2 = max(ans2, w * h);
		}
	printf("%d\n%d\n", ans1, ans2);
	
	return 0;
}