BZOJ 1567 Blue Mary的战役地图(二维hash+二分)

Posted 2020-09-15 freeloop

题意: 求两个矩形最大公共子正方形。(n<=50)

范围这么小可以枚举子正方形的边长。那么可以对这个矩形进行二维hash，就可以在O(1)的时候求出任意子矩形的hash值。然后判断这些正方形的hash值有没有相同的

部分就行了。可以用二分来判断。

需要注意的是行和列乘的hash种子值需要不同的质数，否则可能出现冲突。

时间复杂度O(n^3logn).

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-9
# define MOD 1024523
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    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*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<0) {putchar(‘-‘); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+‘0‘);
}
const int N=55;
//Code begin...

int hash1[N][N], hash2[N][N], P[N*N], Q[N*N], n;
VI a;
VI::iterator it;

bool check(int ans){
    FOR(i,ans,n) FOR(j,ans,n) {
        int tmp=hash2[i][j]-hash2[i-ans][j]*Q[ans]-hash2[i][j-ans]*P[ans]+hash2[i-ans][j-ans]*P[ans]*Q[ans];
        it=lower_bound(a.begin(),a.end(),tmp);
        if (it==a.end()||*it!=tmp) continue;
        return true;
    }
    return false;
}
int main ()
{
    scanf("%d",&n);
    P[0]=1; FOR(i,1,250) P[i]=P[i-1]*131;
    Q[0]=1; FOR(i,1,250) Q[i]=Q[i-1]*1789;
    FOR(i,1,n) FOR(j,1,n) scanf("%d",&hash1[i][j]), hash1[i][j]+=hash1[i][j-1]*P[1];
    FOR(i,1,n) FOR(j,1,n) hash1[i][j]+=hash1[i-1][j]*Q[1];
    FOR(i,1,n) FOR(j,1,n) scanf("%d",&hash2[i][j]), hash2[i][j]+=hash2[i][j-1]*P[1];
    FOR(i,1,n) FOR(j,1,n) hash2[i][j]+=hash2[i-1][j]*Q[1];
    int ans;
    for (ans=n; ans>=1; --ans) {
        a.clear();
        FOR(i,ans,n) FOR(j,ans,n) a.pb(hash1[i][j]-hash1[i-ans][j]*Q[ans]-hash1[i][j-ans]*P[ans]+hash1[i-ans][j-ans]*P[ans]*Q[ans]);
        sort(a.begin(),a.end());
        if (check(ans)) break;
    }
    printf("%d\n",ans);
    return 0;
}

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