CH1806 2351 Matrix

最新推荐文章于 2024-03-17 20:16:04 发布

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#哈希——字符串哈希

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本文介绍了一种使用二维字符串Hash的方法来高效计算矩阵中子矩阵的Hash值，避免了传统取模运算，通过预处理行和列的Hash值，能够在常数时间内计算任意子矩阵的Hash值，用于解决重复子矩阵的问题。

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链接

矩阵

题解

二维字符串 $h a s h$ ，对行和列分别选取一个 $b a s e$ 进行字符串 $h a s h$ ，注意两个 $b a s e$ 不能相同，否则 $A$ 与 $A^T$ 的 $h a s h$ 值就相等了，通过容斥 $O (1)$ 得到子矩阵的 $h a s h$ 值。
这里直接用 $u n s i g n e d$ $l o n g$ $l o n g$ 避开取模。

代码

#include <bits/stdc++.h>

using namespace std;

#define REP(i, n) for (int i = 1; i <= (n); i++)
#define sqr(x) ((x) * (x))

const int maxn = 1000 + 10;
const int maxm = 1000000 + 50;
const int maxt = 500 + 5;

typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;

const LL unit = 1LL;
const int INF = 0x3f3f3f3f;
const LL Inf = 0x3f3f3f3f3f3f3f3fLL;
const double eps = 1e-8;
const double inf = 1e15;
const double pi = acos(-1.0);
const LL mod = 1000000007;

const uLL rp = 131;
const uLL cp = 13331;

int n, m, a, b;
uLL gra[maxn][maxn];
uLL sum[maxn][maxn];
uLL Rpow[maxn], Cpow[maxn];
set<uLL> uset;

inline uLL get_hash(int i, int j, int x, int y, uLL sum[][maxn])
{
    uLL h = sum[x][y];
    h -= sum[i - 1][y] * Rpow[a];
    h -= sum[x][j - 1] * Cpow[b];
    h += sum[i - 1][j - 1] * Rpow[a] * Cpow[b];
    return h;
}

int main()
{
    // ios::sync_with_stdio(false);
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    cin >> n >> m >> a >> b;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            scanf("%1llu", &sum[i][j]);

    Rpow[0] = Cpow[0] = 1;
    for (int i = 1; i <= m; i++)
        Cpow[i] = Cpow[i - 1] * cp;
    for (int i = 1; i <= n; i++)
        Rpow[i] = Rpow[i - 1] * rp;

    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            sum[i][j] += sum[i][j - 1] * cp;

    for (int j = 1; j <= m; j++)
        for (int i = 1; i <= n; i++)
            sum[i][j] += sum[i - 1][j] * rp;

    for (int i = 1; i + a - 1 <= n; i++)
        for (int j = 1; j + b - 1 <= m; j++)
        {
            int x = i + a - 1, y = j + b - 1;
            uset.insert(get_hash(i, j, x, y, sum));
        }

    int q;
    scanf("%d", &q);
    while (q--)
    {
        for (int i = 1; i <= a; i++)
            for (int j = 1; j <= b; j++)
                scanf("%1llu", &gra[i][j]);

        for (int i = 1; i <= a; i++)
            for (int j = 1; j <= b; j++)
                gra[i][j] += gra[i][j - 1] * cp;

        for (int j = 1; j <= b; j++)
            for (int i = 1; i <= a; i++)
                gra[i][j] += gra[i - 1][j] * rp;

        if (uset.count(get_hash(1, 1, a, b, gra)))
            printf("1\n");
        else
            printf("0\n");
    }
    return 0;
}