传送门

f[i][j][0] 表示从右下角到左上角，以（i，j）为起点能延伸的最大值

f[i][j][1] 表示从左下角到右上角，以（i，j）为起点能延伸的最大值

up[i][j] 表示（i，j）上面有多少个0

left[i][j] 表示（i，j）左边有多少个0

right[i][j] 表示（i，j）右边有多少个0

f[i][j][0] = min(f[i - 1][j - 1][0], up[i][j], left[i][j]) + 1

f[i][j][1] = min(f[i - 1][j + 1][1], up[i][j], right[i][j]) + 1

代码

#include <cstdio>
#include <iostream>
#define N 2510
#define max(x, y) ((x) > (y) ? (x) : (y))
#define min(x, y) ((x) < (y) ? (x) : (y))

int n, m, ans;
unsigned short up[N][N], left[N][N], right[N][N], f[N][N][2];
bool a[N][N];

inline int read()
{
    int x = 0, f = 1;
    char ch = getchar();
    for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -1;
    for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - ‘0‘;
    return x * f;
}

int main()
{
	int i, j;
	n = read();
	m = read();
	for(i = 1; i <= n; i++)
		for(j = 1; j <= m; j++)
		{
			f[i][j][0] = f[i][j][1] = a[i][j] = read();
			up[i][j] = a[i - 1][j] ? 0 : up[i - 1][j] + 1;
			left[i][j] = a[i][j - 1] ? 0 : left[i][j - 1] + 1;
		}
	for(i = 1; i <= n; i++)
		for(j = m; j >= 1; j--)
			right[i][j] = a[i][j + 1] ? 0 : right[i][j + 1] + 1;
	for(i = 1; i <= n; i++)
		for(j = 1; j <= m; j++)
		{
			if(!a[i][j]) continue;
			f[i][j][0] = min(f[i - 1][j - 1][0], min(up[i][j], left[i][j])) + 1;
			f[i][j][1] = min(f[i - 1][j + 1][1], min(up[i][j], right[i][j])) + 1;
			ans = max(ans, max(f[i][j][0], f[i][j][1]));
		}
	printf("%d\n", ans);
	return 0;
}