利用FFT来进行字符串匹配

给定串A和串B，A由26个小写字母构成，B由?和26个小写字母构成

?可以和任意字符匹配

求A中出现了多少次B

这里可以使用fft做法，定义向量A和向量B

然后求A和rev(B)的卷积结果C

C的第i-len(B)位就可以表示匹配结果

如果C的第i-len(B)位恰好是B中除了?的字符个数，那么就是匹配成功

这样复杂度就是O((n+m)*(logn + logm))

注意要调整eps，当数据很大的时候，误差会比较大

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <complex>
using namespace std;
const double pi = acos(-);
const int maxn =  ;
typedef complex<double> Complex;
const double eps = 1e-;
void DFT(Complex *a, int n, int t)
{
    if(n == ) return;
    Complex a0[n>>], a1[n>>];
    for(int i = ; i < n; i += ) a0[i>>] = a[i], a1[i>>] = a[i+];
    DFT(a0, n>>, t); DFT(a1, n>>, t);
    Complex wn(cos(*pi/n), t*sin(*pi/n)), w(, );
    for(int i = ; i < (n>>); i++, w *= wn) a[i] = a0[i] + w*a1[i], a[i+(n>>)] = a0[i] - w*a1[i];
}
Complex a[maxn], b[maxn];
int n1, n2, nn, c[maxn];
double x;
string s1, s2;
int main()
{
    freopen("a.txt", "r", stdin);
    cin>>s1>>s2;
    n1 = s1.length(); n2 = s2.length();
    int N = n2;
    for(int i = ; i < n1; i++) x = *pi*(s1[i] - 'a')/, a[i] = Complex(cos(x), sin(x));
    for(int i = ; i < n2; i++)
        if(s2[i] != '?') x = *pi*(s2[i] - 'a')/, b[i] = Complex(cos(-x), sin(-x));
        else b[i] = Complex(, ), N--;
    for(int i = ; i < n2/; i++) swap(b[i], b[n2-i-]);
    n1--; n2--;
    nn = ; while(nn <= n1+n2) nn <<= ;
    DFT(a, nn, ); DFT(b, nn, );
    for(int i = ; i <= nn; i++) a[i] = a[i]*b[i];
    DFT(a, nn, -);
    for(int i = ; i <= n1+n2; i++) c[i] = abs(a[i].imag()) < eps ? (a[i].real()/nn + eps) : ;
    int ans = ;
    for(int i = n2; i <= n1; i++) if(c[i] == N) ans++;
    cout<<ans<<endl;
    return ;
}