我们先处理出两个哈希矩阵，分别代表初始矩阵，转180°后的矩阵。在统计答案时，枚举每个中心点，二分正方形的长度的一半，利用之前处理好的二维哈希值进行O(1)判断。（注意，正方形长度需要分奇偶讨论）
时间复杂度：O(n^2 log n)
#include <bits/stdc++.h>
#define int unsigned long long
using namespace std;
const int N=305;
const int base1=131,base2=83;
int n,m,l,r,mid,ans,sum;
int bin1[N],bin2[N],a[N][N],b[N][N],sum1[N][N],sum2[N][N];
inline bool jay(int x,int y,int len)
{
if (x>n || y>m) return false;
if (x<len || y<len) return false;
int xx=x,yy=y;
int ans1=sum1[x][y]-sum1[x-len][y]*bin1[len]-sum1[x][y-len]*bin2[len]+sum1[x-len][y-len]*bin1[len]*bin2[len];
x=n-(x-len); y=m-(y-len);
int ans2=sum2[x][y]-sum2[x-len][y]*bin1[len]-sum2[x][y-len]*bin2[len]+sum2[x-len][y-len]*bin1[len]*bin2[len];
if (ans1!=ans2) return false;
return true;
}
signed main(){
scanf("%lld%lld",&n,&m);
bin1[0]=bin2[0]=1;
for (register int i=1; i<=n; ++i) bin1[i]=bin1[i-1]*base1;
for (register int i=1; i<=m; ++i) bin2[i]=bin2[i-1]*base2;
for (register int i=1; i<=n; ++i)
for (register int j=1; j<=m; ++j) scanf("%1lld",&a[i][j]);
for (register int i=1; i<=n; ++i)
for (register int j=1; j<=m; ++j) sum1[i][j]=sum1[i][j-1]*base2+a[i][j];
for (register int i=1; i<=n; ++i)
for (register int j=1; j<=m; ++j) sum1[i][j]+=sum1[i-1][j]*base1;
for (register int i=1; i<=n; ++i)
for (register int j=1; j<=m; ++j) b[i][j]=a[n-i+1][m-j+1];
for (register int i=1; i<=n; ++i)
for (register int j=1; j<=m; ++j) sum2[i][j]=sum2[i][j-1]*base2+b[i][j];
for (register int i=1; i<=n; ++i)
for (register int j=1; j<=m; ++j) sum2[i][j]+=sum2[i-1][j]*base1;
for (register int i=1; i<=n; ++i)
for (register int j=1; j<=m; ++j)
{
l=1; r=min(n,m); ans=0;
while (l<=r)
{
mid=l+r>>1;
if (jay(i+mid,j+mid,mid+mid+1)) ans=mid,l=mid+1;
else r=mid-1;
}
sum=max(sum,ans+ans+1);
}
for (register int i=1; i<n; ++i)
for (register int j=1; j<m; ++j)
{
l=1; r=min(n,m); ans=0;
while (l<=r)
{
mid=l+r>>1;
if (jay(i+mid,j+mid,mid+mid)) ans=mid,l=mid+1;
else r=mid-1;
}
sum=max(sum,ans+ans);
}
if (sum>1) printf("%lld\n",sum);
else puts("-1");
return 0;
}