【URAL 1486】Equal Squares（二维哈希+二分）

Description

During a discussion of problems at the Petrozavodsk Training Camp, Vova and Sasha argued about who of them could in 300 minutes find a pair of equal squares of the maximal size in a matrix of size N × M containing lowercase English letters. Squares could overlap each other but could not coincide. He who had found a pair of greater size won. Petr walked by, looked at the matrix, said that the optimal pair of squares had sides K, and walked on. Vova and Sasha still cannot find this pair. Can you help them?

Input

The first line contains integers N and M separated with a space. 1 ≤ N, M ≤ 500. In the next N lines there is a matrix consisting of lowercase English letters, M symbols per line.

Output

In the first line, output the integer K which Petr said. In the next two lines, give coordinates of upper left corners of maximal equal squares. If there exist more than one pair of equal squares of size K, than you may output any of them. The upper left cell of the matrix has coordinates (1, 1), and the lower right cell has coordinates (N, M). If there are no equal squares in the matrix, then output 0.

Sample input

5 10

ljkfghdfas

isdfjksiye

pgljkijlgp

eyisdafdsi

lnpglkfkjl

Sample output

题解

二分答案，然后利用二维哈希\(O(n^2)\)check，总的时间为\(O(n^2 logn)\)

参考代码

#include <map>

#include <queue>

#include <cstdio>

#include <complex>

#include <cstring>

#include <cstdlib>

#include <iostream>

#include <algorithm>

#define ll unsigned long long

#define inf 1000000000

#define PI acos(-1)

#define bug puts("here")

#define REP(i,x,n) for(int i=x;i<=n;i++)

#define DEP(i,n,x) for(int i=n;i>=x;i--)

#define mem(a,x) memset(a,x,sizeof(a))

using namespace std;

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*10+ch-'0';ch=getchar();}

    return x*f;

}

inline void Out(int a){

    if(a<0) putchar('-'),a=-a;

    if(a>=10) Out(a/10);

    putchar(a%10+'0');

}

const int N=505;

char a[N][N];

ll ha[N][N];

ll p1[N],p2[N];

map<ll,int>vis;

int ansx1,ansy1,ansx2,ansy2;

int n,m;

struct node{

   int x,y;

}book[N*N];

bool check(int s){

    vis.clear();

    ll tmp;int tot=0;

    REP(i,s,n) REP(j,s,m){

        tmp=ha[i][j]-ha[i-s][j]*p2[s]-ha[i][j-s]*p1[s]+ha[i-s][j-s]

        *p1[s]*p2[s];

        if(vis[tmp]){

            int id=vis[tmp];

            ansx1=book[id].x;ansy1=book[id].y;

            ansx2=i-s+1;ansy2=j-s+1;

            return true;

        }

        ++tot;

        book[tot].x=i-s+1;book[tot].y=j-s+1;

        vis[tmp]=tot;

    }

    return false;

}

int main(){

    cin>>n>>m;

    REP(i,1,n) cin>>(a[i]+1);

    int seed1=123,seed2=1789;

    REP(i,1,n) REP(j,1,m) ha[i][j]=ha[i][j-1]*seed1+a[i][j];

    REP(i,1,n) REP(j,1,m) ha[i][j]=ha[i-1][j]*seed2+ha[i][j];

    p1[0]=p2[0]=1;

    REP(i,1,m) p1[i]=p1[i-1]*seed1;

    REP(i,1,n) p2[i]=p2[i-1]*seed2;

    int l=1,r=min(n,m),ans=-1;

    while(l<=r){

        int mid=(l+r)>>1;

        if(check(mid)){

            l=mid+1;

            ans=mid;

        }else r=mid-1;

    }

    if(ans!=-1){

        printf("%d\n",ans);

        printf("%d %d\n%d %d\n",ansx1,ansy1,ansx2,ansy2);

    }else puts("0");

    return 0;

}

巴特西