BZOJ 1567: [JSOI2008]Blue Mary的战役地图

传送门

考虑二分答案暴力 $n^4$ 枚举两个矩形右下角，如果能做到 $O(1)$ 判断那么复杂度就可行

那么容易想到二维哈希，然后直接搞就好了...

二维哈希怎么做还是看代码吧...

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
typedef unsigned long long ull;
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
    while(ch>='0'&&ch<='9') { x=(x<<1)+(x<<3)+(ch^48); ch=getchar(); }
    return x*f;
}
const int N=57,base1=9924353,base2=1e9+7;
int n,ans;
ull ha1[N][N],ha2[N][N],hb1[N][N],hb2[N][N],fac1[N],fac2[N];
inline ull get_h(int x,int y,int l,bool flag)
{
    if(flag) return ha2[x][y]-ha2[x][y-l]*fac1[l]-ha2[x-l][y]*fac2[l]+ha2[x-l][y-l]*fac1[l]*fac2[l];
    return hb2[x][y]-hb2[x][y-l]*fac1[l]-hb2[x-l][y]*fac2[l]+hb2[x-l][y-l]*fac1[l]*fac2[l];
}
inline bool check(int p)
{
    for(int i=p;i<=n;i++)
        for(int j=p;j<=n;j++)
            for(int k=p;k<=n;k++)
                for(int l=p;l<=n;l++)
                    if(get_h(i,j,p,1)==get_h(k,l,p,0)) return 1;
    return 0;
}
int main()
{
    n=read();
    fac1[0]=fac2[0]=1; for(int i=1;i<=n;i++) fac1[i]=fac1[i-1]*base1,fac2[i]=fac2[i-1]*base2;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            ha1[i][j]=ha1[i][j-1]*base1+read();
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            hb1[i][j]=hb1[i][j-1]*base1+read();
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            ha2[i][j]=ha2[i-1][j]*base2+ha1[i][j],
            hb2[i][j]=hb2[i-1][j]*base2+hb1[i][j];
        }
    int L=1,R=n;
    while(L<=R)
    {
        int mid=L+R>>1;
        if(check(mid)) L=mid+1,ans=mid;
        else R=mid-1;
    }
    printf("%d\n",ans);
    return 0;
}