正题

题目链接:https://www.luogu.com.cn/problem/P4173

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题目大意

给出两个字符串$S,T$，其中包含小写字母和一些$?$，$?$可以匹配任何字符。

求有多少个$p$使得$T_{0\sim |t|-1}=S_{p\sim p+|t|-1}$

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解题思路

如果不考虑$?$，我们可以用做差法来匹配两个字符，构造匹配函数
$$f(x)=\sum _{i=0}^m(T_i-S_{x+i}{)}^2$$
这样若$f(x)=0$证明它们在位置$x$处匹配。

但是现在有$?$，也就是要跳过有$?$的位置，定义$?$的值为$0$，然后改一下匹配函数$$f(x)=\sum _{i=0}^m(T_i-S_{x+i}{)}^2{T}_i{S}_{x+i}$$

展开二次项$$f(x)=\sum _{i=0}^m{T}_i^3{S}_{x+i}-2T_i^2{S}_{x+i}^2+T_i{S}_{x+i}^3$$
把$T$反过来就是${\sum }_{i=0}^m{T}_{m-i-1}^3{S}_{x+i}-2T_{m-i-1}^2{S}_{x+i}^2+T_{m-i-1}{S}_{x+i}^3$

然后有三个式子卷积之后加起来就好了。

常数极大，开$\text{-O2}$才能过。时间复杂度$O(n\log n)$

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code

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#define int long long
using namespace std;
const int N=1e6+2e5;
const double Pi=acos(-1);
struct complex{double x,y;complex(double xx=0,double yy=0){x=xx;y=yy;return;}
};
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);}
struct poly{complex a[N];
}F[3],G[3];
int n,m,r[N],k;
char s[N],t[N];
queue<int> q;
void FFT(complex *f,int op,int n){for(int i=0;i<n;i++)if(i<r[i])swap(f[i],f[r[i]]);for(int p=2;p<=n;p<<=1){int len=p>>1;complex tmp(cos(Pi/len),sin(Pi/len)*op);for(int k=0;k<n;k+=p){complex buf(1,0);for(int i=k;i<k+len;i++){complex tt=buf*f[i+len];f[i+len]=f[i]-tt;f[i]=f[i]+tt;buf=buf*tmp;}}}if(op==-1)for(int i=0;i<n;i++)f[i].x/=(double)n;return;
}
void mul(poly &a,poly &b){FFT(a.a,1,k);FFT(b.a,1,k);for(int i=0;i<k;i++)a.a[i]=a.a[i]*b.a[i];FFT(a.a,-1,k);return;
}
signed main()
{scanf("%d%d",&m,&n);scanf("%s",t);scanf("%s",s);for(int i=0;i<n;i++){char c=s[i];int z=s[i]-'a'+1;if(c=='*')z=0;F[0].a[i]=complex(z*z*z,0);F[1].a[i]=complex(z*z,0);F[2].a[i]=complex(z,0);}for(int i=0;i<m;i++){char c=t[m-i-1];int z=c-'a'+1;if(c=='*')z=0;G[0].a[i]=complex(z,0);G[1].a[i]=complex(z*z,0);G[2].a[i]=complex(z*z*z,0);}k=1;for(k=1;k<=2*n;)k<<=1;for(int i=0;i<k;i++)r[i]=(r[i>>1]>>1)|((i&1)?(k>>1):0);mul(F[0],G[0]);mul(F[1],G[1]);mul(F[2],G[2]);for(int i=0;i<k;i++)F[0].a[i].x=F[0].a[i].x-2.0*F[1].a[i].x+F[2].a[i].x;k=0;for(int i=0;i<=n-m;i++)if(fabs(F[0].a[i+m-1].x)<0.5)k++,q.push(i);printf("%d\n",k);while(!q.empty())printf("%d ",q.front()+1),q.pop();return 0;
}