题目

https://ac.nowcoder.com/acm/contest/2?&headNav=www#question

解析 我们对矩阵进行二维hash，所以每个子矩阵都有一个额hash值，二分答案然后O(n^2) check 枚举矩阵终点，记录每个hash值与有两个一样的就true

AC代码

#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
const int maxn=2e6+10;
const ull base1=131,base2=233; //base，基数

int n, m;
char mp[510][510];
ull has[510][510];
ull p1[510], p2[510];
map<ull, int> mmp;

void init()
{
    p1[0] = p2[0] = 1;
    for(int i = 1; i <= 505; i ++)
    {
        p1[i] = p1[i-1]*base1;
        p2[i] = p2[i-1]*base2;
    }
}

void Hash()
{
    has[0][0] = 0;
    has[0][1] = 0;
    has[1][0] = 0;
    for(int i = 1; i <= n; i ++)
    {
        for(int j = 1; j <= m; j ++)
        {
            has[i][j] = has[i][j-1]*base1 + mp[i][j] - 'a';
        }
    }
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j ++)
        {
            has[i][j] = has[i-1][j]*base2 + has[i][j];
        }
    }
}

bool check(int x)
{
    mmp.clear();
    for(int i = x; i <= n; i ++)
    {
        for(int j = x; j <= m; j ++)
        {
            ull k = has[i][j] - has[i-x][j]*p2[x] - has[i][j-x]*p1[x] + has[i-x][j-x]*p1[x]*p2[x];
            mmp[k] ++;
            if(mmp[k] >= 2)
                return true;
        }
    }
    return false;
}

int main()
{
    init();
    while(~scanf("%d%d", &n, &m))
    {
        for(int i = 1; i <= n; i ++)
        {
            scanf("%s", mp[i]+1);
        }
        Hash();
        int l = 0, r = 600, ans = 0;
        while(l <= r)
        {
            int mid = (l+r)/2;
            if(check(mid))
            {
                l = mid+1;
                ans = mid;
            }
            else
            {
                r = mid-1;
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}