牛客练习赛1 矩阵 字符串二维hash+二分

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+;
const ull base1=,base2=; //base，基数

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

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

void Hash()
{
    has[][] = ;
    has[][] = ;
    has[][] = ;
    for(int i = ; i <= n; i ++)
    {
        for(int j = ; j <= m; j ++)
        {
            has[i][j] = has[i][j-]*base1 + mp[i][j] - 'a';
        }
    }
    for(int i = ; i <= n; i++)
    {
        for(int j = ; j <= m; j ++)
        {
            has[i][j] = has[i-][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] >= )
                return true;
        }
    }
    return false;
}

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