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;
}

BZOJ 1567 JSOI2008 Blue Mary的战役地图 Hash+二分