BZOJ4259: 殘缺的字元串(FFT 字元串匹配)

題意

題目鏈接

Sol

知道FFT能做字元串匹配的話這就是個裸題了吧。。

考慮把B翻轉過來，如果\(\sum_{k = 0}^M (B_{i - k} - A_k)^2 * B_{i-k}*A_k = 0\)

那麼說明能匹配。然後拆開三波FFT就行了

/*

*/
#include<bits/stdc++.h>
#define LL long long 
const int MAXN = 1e6 + 10, INF = 1e9 + 7;
using namespace std;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M;
LL g[MAXN], f[MAXN];
char sa[MAXN], sb[MAXN];
int ta[MAXN], tb[MAXN], a[MAXN], b[MAXN], rev[MAXN], lim;
LL sqr2(int x) {return 1ll * x * x;}
LL sqr3(int x) {return 1ll * x * x * x;}
const double PI = acos(-1);
struct com {
    double x, y;
    com operator * (const com &rhs) const {
        return {x * rhs.x - y * rhs.y, x * rhs.y + y * rhs.x};
    }
    com operator + (const com &rhs) const {
        return {x + rhs.x, y + rhs.y};
    }
    com operator - (const com &rhs) const {
        return {x - rhs.x, y - rhs.y};
    }
}A[MAXN], B[MAXN];
void FFT(com *A, int lim, int type) {
    for(int i = 0; i < lim; i++) if(i < rev[i]) swap(A[i], A[rev[i]]);
    for(int mid = 1; mid < lim; mid <<= 1) {
        com wn = {cos(PI / mid), type * sin(PI / mid)};
        for(int i = 0; i < lim; i += (mid << 1)) {
            com w = {1, 0}; 
            for(int j = 0; j < mid; j++, w = w * wn) {
                com x = A[i + j], y = w * A[i + j + mid];
                A[i + j] = x + y;
                A[i + j + mid] = x - y;
            }
        }
    }
    if(type == -1) {
        for(int i = 0; i <= lim; i++) A[i].x /= lim;
    }
}
void mul(int *b, int *a) {
    memset(A, 0, sizeof(A)); memset(B, 0, sizeof(B));
    for(int i = 0; i < N; i++) B[i].x = b[i];
    for(int i = 0; i < M; i++) A[i].x = a[i];
    FFT(B, lim, 1);
    FFT(A, lim, 1);
    for(int i = 0; i < lim; i++) B[i] = B[i] * A[i];
    FFT(B, lim, -1);
    for(int i = M - 1; i <= N; i++) 
        f[i] += round(B[i].x);
}
signed main() {
    //freopen("2.in", "r", stdin);  freopen("b.out", "w", stdout);
    M = read(); N = read();
    scanf("%s %s", sa, sb);
    for(int i = 0; i < M; i++) ta[i] = (sa[i] == '*' ? 0 : sa[i] - 'a' + 1);
    for(int i = 0; i < N; i++) tb[i] = (sb[i] == '*' ? 0 : sb[i] - 'a' + 1);
    reverse(tb, tb + N);

    int len = 0; lim = 1;
    while(lim <= N + M) len++, lim <<= 1;
    for(int i = 0; i < lim; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << len - 1);
    
    for(int i = 0; i < N; i++) b[i] = sqr3(tb[i]);
    for(int i = 0; i < M; i++) a[i] = ta[i]; 
    mul(b, a);
    
    for(int i = 0; i < N; i++) b[i] = -2 * sqr2(tb[i]);
    for(int i = 0; i < M; i++) a[i] = sqr2(ta[i]);
    mul(b, a);

    for(int i = 0; i < N; i++) b[i] = tb[i];
    for(int i = 0; i < M; i++) a[i] = sqr3(ta[i]);
    mul(b, a);  
    
    int ans = 0;
    for(int i = M - 1; i < N; i++) 
        if(!f[i]) ans++;

    printf("%d\n", ans);

    for(int i = N - 1; i >= M - 1; i--) 
        if(!f[i]) 
            printf("%d ", N - i);
    
    return 0;
}
/*
3 7
a*b
aebr*ob
*/