CF954I Yet Another String Matching Problem

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给定两个字符串𝑆,𝑇

求𝑆所有长度为|𝑇|子串与𝑇的距离

两个等长的串的距离定义为最少的，将某一个字符全部视作另外一个字符的次数。

$|𝑇|\leq|𝑆|\leq 10^6$，字符集大小为6

$abac$与$aacb$，其实只$a-b,c-a,b-c$，对于一个联通块只需要改$size-1$即可，字符集只有$6$,只用维护$6\times 6$，是否在匹配时有$ab,ac$这样子。用并差集维护即可。

代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 5e5 + 10;
const int mod = 1e9 + 7;
const int P = 998244353, gi = 3;
const double pi = acos(-1.0);
int inc(int x, int y, int mo)
{
    if (x + y >= mo)
        return (x + y - mo);
    else
        return (x + y);
}

int del(int x, int y, int mo)
{
    if (x - y < 0)
        return (x - y + mo);
    else
        return (x - y);
}
int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}

int qpow(int a, int x, int mo)
{
    int res = 1;
    while (x)
    {
        if (x & 1)
            res = 1ll * res * a % mo;
        x >>= 1;
        a = 1ll * a * a % mo;
    }
    return res;
}
int rev[N];
void NTT(int *A, int n, int inv)
{
    for (int i = 0; i < n; i++)
        if (i < rev[i])
            swap(A[i], A[rev[i]]);
    for (int l = 1; l < n; l <<= 1)
    {
        int tt = qpow(gi, (P - 1) / (l << 1), P);
        int temp = (inv == 1 ? tt : qpow(tt, P - 2, P));
        for (int i = 0; i < n; i += (l << 1))
        {
            int omega = 1;
            for (int j = 0; j < l; j++, omega = 1ll * omega * temp % P)
            {
                int x = A[i + j], y = 1ll * omega * A[i + j + l] % P;
                A[i + j] = inc(x, y, P);
                A[i + j + l] = del(x, y, P);
            }
        }
    }
    int invv = qpow(n, P - 2, P);
    if (inv == -1)
        for (int i = 0; i < n; i++)
            A[i] = 1ll * A[i] * invv % P;
}

void NTTX(int *a, int n, int *b, int m)
{
    int ML = 1, bit = 0;
    while (ML < n + m)
        ML <<= 1, bit++;
    for (int i = 0; i < ML; i++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    NTT(a, ML, 1);
    NTT(b, ML, 1);
    for (int i = 0; i < ML; i++)
        a[i] = 1ll * a[i] * b[i] % P;
    NTT(a, ML, -1);
}
int slen, tlen;
int fa[N][6];
int f[N], g[N];
char s[N], t[N];
int ans[N];
int find(int p, int x)
{
    if (fa[p][x] != x)
        return fa[p][x] = find(p, fa[p][x]);
    return x;
}
void solve(int x, int y)
{

    memset(f, 0, sizeof(f));
    memset(g, 0, sizeof(g));
    for (int i = 1; i <= slen; i++)
        f[i] = (s[i] - 'a') == x;
    for (int i = 1; i <= tlen; i++)
        g[i] = (t[i] - 'a') == y;
    reverse(f + 1, f + slen + 1);

    NTTX(f, slen, g, tlen);
    for (int p = 1; p <= slen - tlen + 1; p++)
    {
        if (f[slen - p + 2] > 0)
        {

            int xx = find(p, x);
            int yy = find(p, y);
            if (xx != yy)
                fa[p][xx] = yy, ans[p]++;
        }
    }
}
int main()
{
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    slen = strlen(s + 1);
    tlen = strlen(t + 1);
    for (int i = 1; i < N; i++)
        for (int j = 0; j < 6; j++)
            fa[i][j] = j;
    for (int i = 0; i < 6; i++)
        for (int j = 0; j < 6; j++)
            if (i != j)
                solve(i, j);
    for (int p = 1; p <= slen - tlen + 1; p++)
        printf("%d ", ans[p]);
}