題目描述

傳送門

題目大意：給出一個模板串和一個母串，問模板串在母串中出現過幾次。帶萬用字元。

題解

這道題和兩個串那道題是差不多的。。
令F(i)表示將模板串的最後一個懟到母串的第i個是否能匹配，0表示能匹配，非0表示不能匹配。然後設兩個函式f(i)=(t(i)=‘ * ’)?0:t(i),g(i)=(s(i)=‘ * ’)?0:s(i)
將模板串倒置了之後顯然F(i)=∑j=0S−1f(i−j)g(j)(f(i−j)−g(j))2=f(i−j)3g(j)−2f(i−j)2g(j)2+f(i−j)g(j)3是一個卷積的形式
這樣做三遍FFT求出F(i)就行了

程式碼

#include<algorithm>
 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define N 2000005

const double pi=acos(-1.0);
struct complex
{
    double x,y;
    complex(double X=0,double Y=0)
    {
        x=X,y=Y;
    }
}a[N],b[N];
complex operator + (complex a,complex
 
 b) {return complex(a.x+b.x,a.y+b.y);}
complex operator - (complex a,complex b) {return complex(a.x-b.x,a.y-b.y);}
complex operator * (complex a,complex b) {return complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
int S,T,n,m,L,R[N],ans[N];
long long F[N];
char s[N],t[N];
double f[N],g[N];

void FFT(complex
 
 a[N],int opt)
{
    for (int i=0;i<n;++i)
        if (i<R[i]) swap(a[i],a[R[i]]);
    for (int k=1;k<n;k<<=1)
    {
        complex wn=complex(cos(pi/k),opt*sin(pi/k));
        for (int i=0;i<n;i+=(k<<1))
        {
            complex w=complex(1,0);
            for (int j=0;j<k;++j,w=w*wn)
            {
                complex x=a[i+j],y=w*a[i+j+k];
                a[i+j]=x+y,a[i+j+k]=x-y;
            }
        }
    }
}
void calc(int opt)
{
    FFT(a,1);FFT(b,1);
    for (int i=0;i<=n;++i) a[i]=a[i]*b[i];
    FFT(a,-1);
    for (int i=0;i<T;++i)
        F[i]+=(long long)(a[i].x/n+0.5)*opt;
}
int main()
{
    scanf("%d%d",&S,&T);
    scanf("%s%s",s,t);
    for (int i=0;i<S/2;++i) swap(s[i],s[S-i-1]);

    for (int i=0;i<T;++i) f[i]=(t[i]=='*')?0:(t[i]-'a'+1.0);
    for (int i=0;i<S;++i) g[i]=(s[i]=='*')?0:(s[i]-'a'+1.0);

    m=S+T-2;
    for (n=1;n<=m;n<<=1) ++L;
    for (int i=0;i<n;++i)
        R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));

    for (int i=0;i<=n;++i) a[i]=complex(0,0),b[i]=complex(0,0);
    for (int i=0;i<T;++i) a[i].x=f[i]*f[i]*f[i];
    for (int i=0;i<S;++i) b[i].x=g[i];
    calc(1);
    for (int i=0;i<=n;++i) a[i]=complex(0,0),b[i]=complex(0,0);
    for (int i=0;i<T;++i) a[i].x=f[i]*f[i];
    for (int i=0;i<S;++i) b[i].x=g[i]*g[i];
    calc(-2);
    for (int i=0;i<=n;++i) a[i]=complex(0,0),b[i]=complex(0,0);
    for (int i=0;i<T;++i) a[i].x=f[i];
    for (int i=0;i<S;++i) b[i].x=g[i]*g[i]*g[i];
    calc(1);

    for (int i=S-1;i<T;++i)
        if (!F[i]) ans[++ans[0]]=i-S+2;
    printf("%d\n",ans[0]);
    for (int i=1;i<=ans[0];++i) printf("%d%c",ans[i]," \n"[i==ans[0]]);
}