BZOJ1057或洛谷1169 [ZJOI2007]棋盘制作

BZOJ原题链接

洛谷原题链接

设\(L[i][j],R[i][j],H[i][j]\)表示点\((i,j)\)向左、右、上尽量拓展的左端点、右端点、上端点的坐标。

\(L,R\)直接初始化好，\(H\)则全部为\(1\)。

扫过整个矩阵，对于每个点，尽量去拓展上端点，并更新\(L[i][j] = \max\{ L[i][j], L[i - 1][j] \}, R[i][j] = \min\{ R[i][j], R[i - 1][j] \}, H[i][j] = H[i - 1][j] + 1\)，然后尝试去更新答案即可。

不得不吐糟一句，某谷的数据是真的水。

#include<cstdio>
using namespace std;
const int N = 2010;
int a[N][N], L[N][N], R[N][N], H[N][N];
inline int re()
{
	int x = 0;
	char c = getchar();
	bool p = 0;
	for (; c < '0' || c > '9'; c = getchar())
		p |= c == '-';
	for (; c >= '0' && c <= '9'; c = getchar())
		x = x * 10 + c - '0';
	return p ? -x : x;
}
inline int maxn(int x, int y)
{
	return x > y ? x : y;
}
inline int minn(int x, int y)
{
	return x < y ? x : y;
}
inline int squ(int x)
{
	return x * x;
}
int main()
{
	int i, j, n, m, l, sq = 1, rec = 1;
	n = re();
	m = re();
	for (i = 1; i <= n; i++)
		for (j = 1; j <= m; j++)
		{
			a[i][j] = re();
			L[i][j] = R[i][j] = j;
			H[i][j] = 1;
		}
	for (i = 1; i <= n; i++)
		for (j = 2; j <= m; j++)
			if (a[i][j] ^ a[i][j - 1])
				L[i][j] = L[i][j - 1];
	for (i = 1; i <= n; i++)
		for (j = m - 1; j; j--)
			if (a[i][j] ^ a[i][j + 1])
				R[i][j] = R[i][j + 1];
	for (i = 1; i <= n; i++)
		for (j = 1; j <= m; j++)
		{
			if (i > 1 && a[i][j] ^ a[i - 1][j])
			{
				L[i][j] = maxn(L[i][j], L[i - 1][j]);
				R[i][j] = minn(R[i][j], R[i - 1][j]);
				H[i][j] = H[i - 1][j] + 1;
			}
			l = R[i][j] - L[i][j] + 1;
			sq = maxn(sq, squ(minn(l, H[i][j])));
			rec = maxn(rec, l * H[i][j]);
		}
	printf("%d\n%d", sq, rec);
	return 0;
}