题意

给你两个矩阵，问你最大的相同正方形的边长是多少

题解

二维hash
二分答案
暴力检测就好了
hash方法用的是字符串的QAQ
应该是通用的吧。。

#include<cstdio>
#include<cstring>
#include<cstring>
typedef long long LL;
const LL MOD=10037;//行
const LL MOD1=100003;//列 
const LL MOD2=1000000007; //总共 
const LL N=55;
LL n;
LL a[N][N];
LL f[N][N];//hash的矩阵
LL f1[N][N];
LL G[N],G1[N];
LL Get (LL x,LL y,LL z)//以x,y为左上角 
{LL c=x+z-1,d=y+z-1;LL ans=(f[c][d]-f[x-1][d]*G1[z]%MOD2)%MOD2;ans=(ans-f[c][y-1]*G[z]%MOD2)%MOD2;ans=(ans+f[x-1][y-1]*G[z]%MOD2*G1[z]%MOD2)%MOD2;ans=(ans+MOD2)%MOD2;return ans;
}
LL Get1 (LL x,LL y,LL z)//以x,y为左上角 
{LL c=x+z-1,d=y+z-1;LL ans=(f1[c][d]-f1[x-1][d]*G1[z]%MOD2)%MOD2;ans=(ans-f1[c][y-1]*G[z]%MOD2)%MOD2;ans=(ans+f1[x-1][y-1]*G[z]%MOD2*G1[z]%MOD2)%MOD2;ans=(ans+MOD2)%MOD2;return ans;
}
void Ins ()
{scanf("%lld",&n);for (LL u=1;u<=n;u++)for (LL i=1;i<=n;i++)scanf("%lld",&a[u][i]);for (LL u=1;u<=n;u++)for (LL i=1;i<=n;i++)f[u][i]=(f[u][i-1]*MOD+a[u][i])%MOD2;for (LL u=1;u<=n;u++)for (LL i=1;i<=n;i++)f[i][u]=(f[i-1][u]*MOD1+f[i][u])%MOD2;for (LL u=1;u<=n;u++)for (LL i=1;i<=n;i++)scanf("%lld",&a[u][i]);for (LL u=1;u<=n;u++)for (LL i=1;i<=n;i++)f1[u][i]=(f1[u][i-1]*MOD+a[u][i])%MOD2;for (LL u=1;u<=n;u++)for (LL i=1;i<=n;i++)f1[i][u]=(f1[i-1][u]*MOD1+f1[i][u])%MOD2;
}
bool check (LL u)
{for (LL i=1;i<=n-u+1;i++)for (LL j=1;j<=n-u+1;j++)for (LL k=1;k<=n-u+1;k++)for (LL l=1;l<=n-u+1;l++)if (Get(i,j,u)==Get1(k,l,u))return true;return false;
}
int main()
{Ins();G[0]=1;for (LL u=1;u<=n;u++) G[u]=(G[u-1]*MOD)%MOD2;G1[0]=1;for (LL u=1;u<=n;u++) G1[u]=(G1[u-1]*MOD1)%MOD2;LL l=1,r=n;LL ans=0;while (l<=r){LL mid=(l+r)>>1;if (check(mid)) {ans=mid;l=mid+1;}else r=mid-1;}printf("%lld\n",ans);return 0;
}