Description

定義兩個長度相等的字串之間的距離為：

把兩個字串中所有同一種字元變成另外一種，使得兩個字串相等所需要操作的次數的最小值。

求 $s$中每一個長度為 $\mathrm{\mid }t\mathrm{\mid }$

字符集為小寫字母a到f。

Solution

考慮如果只有兩個串怎麼做，肯定是從左到右掃一遍，如果某兩個字元不相同，則用並查集判斷是否在同一個集合裡，如果不在則加入同一個集合並且答案加 $1$。

考慮列舉兩個不同的字元看它們是否連邊。列舉兩個不同的字元，將 $S$

假設在 $S$的位置 $a$與 $T$的位置 $b$均為 $1$。那麼意味著在兩串末尾匹配到 $a+|T|-b$時，字元 $x$向 $y$連了一條邊，將 $T$反轉，上式轉化為卷積的形式，FFT即可。

#include <bits/stdc++.h>
using namespace std;

const int maxn = 125005 * 5;

int n, m;
char s[maxn], t[maxn];
int v1[maxn], v2[maxn], f[maxn][6], ans[maxn];

inline int gi()
{
    char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    int sum = 0;
    while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
    return sum;
}

namespace FFT
{

    const double pi = acos(-1);
    typedef complex<double> cpx;
    
    int n, len, L, R[maxn];
    cpx A[maxn], B[maxn];

    void FFT(cpx *a, int f)
    {
        for (int i = 0; i < n; ++i) if (i < R[i]) swap(a[i], a[R[i]]);
        for (int i = 1; i < n; i <<= 1) {
            cpx wn(cos(pi / i), sin(f * pi / i)), t;
            for (int j = 0; j < n; j += (i << 1)) {
                cpx w(1, 0);
                for (int k = 0; k < i; ++k, w = w * wn) {
                    t = a[j + i + k] * w;
                    a[j + i + k] = a[j + k] - t;
                    a[j + k] = a[j + k] + t;
                }
            }
        }
    }
    
    void mul(int *a, int *b, int len1, int len2, int *c)
    {
        L = 0;
        for (n = 1, len = len1 + len2 - 1; n < len; n <<= 1) ++L;
        --L;
        for (int i = 0; i < n; ++i) R[i] = (R[i >> 1] >> 1) | ((i & 1) << L);

        fill(A, A + n, 0); fill(B, B + n, 0);
        for (int i = 0; i < len1; ++i) A[i] = a[i];
        for (int i = 0; i < len2; ++i) B[i] = b[i];
        FFT(A, 1); FFT(B, 1);
        for (int i = 0; i < n; ++i) A[i] *= B[i];
        FFT(A, -1);

        for (int i = 0; i < len1; ++i) c[i] = (int)(A[i].real() / n + 0.5);
    }

}

inline int find(int *f, int x)
{
    if (f[x] == x) return x;
    return f[x] = find(f, f[x]);
}

void solve(int c1, int c2)
{
    for (int i = 0; i < n; ++i) v1[i] = s[i] == 'a' + c1;
    for (int i = 0; i < m; ++i) v2[i] = t[i] == 'a' + c2;
    reverse(v2, v2 + m);

    FFT::mul(v1, v2, n, m, v1);

    for (int i = 0; i < n; ++i) {
        if (!v1[i]) continue;
        if (find(f[i], c1) != find(f[i], c2)) f[i][f[i][c1]] = f[i][c2], ++ans[i];
    }
}

int main()
{
    scanf("%s\n", s); n = strlen(s);
    scanf("%s\n", t); m = strlen(t);
    
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < 6; ++j) f[i][j] = j;

    for (int i = 0; i < 6; ++i)
        for (int j = 0; j < 6; ++j) {
            if (i == j) continue;
            solve(i, j);
        }
            
    for (int i = m - 1; i < n; ++i) printf("%d ", ans[i]);
    puts("");
    
    return 0;
}