[BZOJ1567][JSOI2008]Blue Mary的战役地图（二分+矩阵hash）

题目描述

传送门

题解

二分答案之后 O(n2) 矩阵hash
就是个裸题

代码

#include
#include
#include
#include
#include
#include
using namespace std;
#define UL unsigned long long
#define N 55

const UL s=23333333333LL;
const UL S=3115439631056497LL;
int n,ans,a[N][N],b[N][N];
UL mis[N],miS[N],c[N][N],d[N][N];
map bool> hash;

bool check(int mid)
{
    for (int i=1;i<=n;++i)
    {
        c[i][1]=(UL)a[i][1];
        for (int j=2;j<=mid;++j)
            c[i][1]=c[i][1]*s+(UL)a[i][j];
        for (int j=2;j<=n-mid+1;++j)
            c[i][j]=(c[i][j-1]-(UL)a[i][j-1]*mis[mid-1])*s+(UL)a[i][j+mid-1];
    }
    for (int i=1;i<=n-mid+1;++i)
    {
        d[1][i]=c[1][i];
        for (int j=2;j<=mid;++j)
            d[1][i]=d[1][i]*S+c[j][i];
        for (int j=2;j<=n-mid+1;++j)
            d[j][i]=(d[j-1][i]-c[j-1][i]*miS[mid-1])*S+c[j+mid-1][i];
    }
    hash.clear();
    for (int i=1;i<=n-mid+1;++i)
        for (int j=1;j<=n-mid+1;++j)
            hash[d[i][j]]=1;

    for (int i=1;i<=n;++i)
    {
        c[i][1]=(UL)b[i][1];
        for (int j=2;j<=mid;++j)
            c[i][1]=c[i][1]*s+(UL)b[i][j];
        for (int j=2;j<=n-mid+1;++j)
            c[i][j]=(c[i][j-1]-(UL)b[i][j-1]*mis[mid-1])*s+(UL)b[i][j+mid-1];
    }
    for (int i=1;i<=n-mid+1;++i)
    {
        d[1][i]=c[1][i];
        for (int j=2;j<=mid;++j)
            d[1][i]=d[1][i]*S+c[j][i];
        for (int j=2;j<=n-mid+1;++j)
            d[j][i]=(d[j-1][i]-c[j-1][i]*miS[mid-1])*S+c[j+mid-1][i];
    }
    for (int i=1;i<=n-mid+1;++i)
        for (int j=1;j<=n-mid+1;++j)
            if (hash[d[i][j]]) return 1;
    return 0;
}
int find()
{
    int l=1,r=n,mid,ans=0;
    while (l<=r)
    {
        mid=(l+r)>>1;
        if (check(mid)) ans=mid,l=mid+1;
        else r=mid-1;
    }
    return ans;
}
int main()
{
    scanf("%d",&n);
    mis[0]=1LL;for (int i=1;i<=n;++i) mis[i]=mis[i-1]*s;
    miS[0]=1LL;for (int i=1;i<=n;++i) miS[i]=miS[i-1]*S;
    for (int i=1;i<=n;++i)
        for (int j=1;j<=n;++j)
            scanf("%d",&a[i][j]);
    for (int i=1;i<=n;++i)
        for (int j=1;j<=n;++j)
            scanf("%d",&b[i][j]);
    ans=find();
    printf("%d\n",ans);
}