[BZOJ4259]残缺的字符串（FFT）

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题目描述

传送门

题目大意：给出一个模板串和一个母串，问模板串在母串中出现过几次。带通配符。

题解

这道题和两个串那道题是差不多的。。
令F(i)表示将模板串的最后一个怼到母串的第i个是否能匹配，0表示能匹配，非0表示不能匹配。然后设两个函数f(i)=(t(i)=‘ * ’)?0:t(i),g(i)=(s(i)=‘ * ’)?0:s(i)
将模板串倒置了之后显然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是一个卷积的形式
这样做三遍FFT求出F(i)就行了

代码

#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<cmath>using namespace std;#define N 2000005const 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]]);}

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