题目大意：给定两个矩阵，求最大公共子正方形边长
首先二分答案 然后Check的时候先把A矩阵的所有边长为x的子正方形存在哈希表里 然后枚举B矩阵的每个子正方形查找
注意二维哈希的时候横竖用的两个BASE不能一样 否则当两个矩阵关于对角线对称的时候会判断为相等
尼玛我的哈希表居然比map慢……不活了
#include<map>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define M 60
#define SIZE 1001001
#define BASE1 999911657
#define BASE2 999911659
using namespace std;
typedef unsigned int u;
int n,T;
u a[M][M],b[M][M],power1[M],power2[M];
map<u,int>table;
bool Judge(int x)
{
int i,j;
++T;
for(i=x;i<=n;i++)
for(j=x;j<=n;j++)
{
u hash=a[i][j]
-a[i-x][j]*power1[x]
-a[i][j-x]*power2[x]
+a[i-x][j-x]*power1[x]*power2[x];
table[hash]=T;
}
for(i=x;i<=n;i++)
for(j=x;j<=n;j++)
{
u hash=b[i][j]
-b[i-x][j]*power1[x]
-b[i][j-x]*power2[x]
+b[i-x][j-x]*power1[x]*power2[x];
if( table[hash]==T )
return true;
}
return false;
}
int Bisection()
{
int l=0,r=n;
while(l+1<r)
{
int mid=l+r>>1;
if( Judge(mid) )
l=mid;
else
r=mid;
}
if( Judge(r) )
return r;
return l;
}
int main()
{
int i,j;
cin>>n;
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
scanf("%u",&a[i][j]);
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
a[i][j]+=a[i-1][j]*BASE1;
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
a[i][j]+=a[i][j-1]*BASE2;
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
scanf("%u",&b[i][j]);
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
b[i][j]+=b[i-1][j]*BASE1;
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
b[i][j]+=b[i][j-1]*BASE2;
power1[0]=power2[0]=1;
for(i=1;i<=n;i++)
power1[i]=power1[i-1]*BASE1,
power2[i]=power2[i-1]*BASE2;
cout<<Bisection()<<endl;
}
