《ybtoj高效进阶》第二部分第二章例题5 子正方形

题目大意

给2个正方形，求最大公共子正方形的边长。
2个正方形边长<=50

思路

显然可以通过二维hash+二分答案+暴力枚举达到目的
code:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
using namespace std;
int a[1001][1001],b[1001][1001],ans;
int n,m;
unsigned long long ha[1001][1001],hb[1001][1001],p[1001],p2[1001];//数组开大又不会死
bool check(int x,int y,int l)
{
    
    
	if (x>n||y>m||x<l||y<l) return 0;
	unsigned long long aa,bb;
	aa=ha[x][y]-ha[x][y-l]*p[l]-ha[x-l][y]*p2[l]+ha[x-l][y-l]*p[l]*p2[l];
	for (int xx=l;xx<=n;xx++)
	{
    
    
		for (int yy=l;yy<=m;yy++)
		{
    
    
			if (xx>n||yy>m) continue;
			bb=hb[xx][yy]-hb[xx][yy-l]*p[l]-hb[xx-l][yy]*p2[l]+hb[xx-l][yy-l]*p[l]*p2[l];
			if (bb==aa) return 1;
		}
	}
	return 0;
}
int main()
{
    
    
	cin>>n;
	m=n;
	for (int i=1;i<=n;i++)
	{
    
    
		for (int j=1;j<=m;j++)
		{
    
    
			cin>>a[i][j];
		}
	}
	for (int i=1;i<=n;i++)
	{
    
    
		for (int j=1;j<=m;j++)
		{
    
    
			cin>>b[i][j];
		}
	}
	p2[0]=p[0]=1;
	for (int i=1;i<=1000;i++) p[i]=97*p[i-1],p2[i]=13331*p2[i-1];
	for (int i=1;i<=n;i++)
	{
    
    
		for (int j=1;j<=m;j++)
		{
    
    
			ha[i][j]=ha[i][j-1]*97+a[i][j];
			hb[i][j]=hb[i][j-1]*97+b[i][j];
		}
	}
	for (int i=2;i<=n;i++)
	{
    
    
		for (int j=1;j<=m;j++)
		{
    
    
			ha[i][j]=ha[i-1][j]*13331+ha[i][j];
			hb[i][j]=hb[i-1][j]*13331+hb[i][j];
		}
	}
	for (int i=1;i<=n;i++)
	{
    
    
		for (int j=1;j<=m;j++)
		{
    
    
			int l=1,r=n+m,mn=0;
			while (l<=r)
			{
    
    
				int mid=(l+r)>>1;
				if (check(i+mid,j+mid,mid*2+1)) l=mid+1,mn=max(mn,mid*2+1);
				else r=mid-1;
			}
			ans=max(ans,mn);
			l=1,r=n+m,mn=0;
			while (l<=r)
			{
    
    
				int mid=(l+r)>>1;
				if (check(i+mid,j+mid,mid*2)) l=mid+1,mn=max(mn,mid*2);
				else r=mid-1;
			}
			ans=max(ans,mn);
		}
	}
	cout<<ans;
    return 0;
}