Description:

给定两个带通配符的串,求可能出现几次匹配,以及这些匹配位置

Hint:

\(n \le 3*10^5\)

Solution:

定义匹配函数 \(P(x)=\sum_{i=x}^{x+m}(S1[i]-S2[i])^2*S1[i]*S2[i]​\)

展开的式子太长,有时间再放

大概是一堆字符串卷积

翻转后FFT即可

#include <map>

#include <set>

#include <stack>

#include <cmath>

#include <queue>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <iostream>

#include <algorithm>

#define ls p<<1

#define rs p<<1|1

using namespace std;

typedef long long ll;

const int mxn=2e6+5;

const double PI=acos(-1);

int n,m,l,tot,lim=1,a[mxn],b[mxn],r[mxn],q[mxn];

ll s[mxn];

char s1[mxn],s2[mxn];

inline int read() {

    char c=getchar(); int x=0,f=1;

    while(c>'9'||c<'0') {if(c=='-') f=-1;c=getchar();}

    while(c<='9'&&c>='0') {x=(x<<3)+(x<<1)+(c&15);c=getchar();}

    return x*f;

}

inline int chkmax(register int &x,register int y) {if(x<y) x=y;}

inline int chkmin(register int &x,register int y) {if(x>y) x=y;}

struct ed {

    int to,nxt;

}t[mxn<<1];

struct cp {

    double x,y;

    cp (double xx=0,double yy=0) {x=xx;y=yy;}

    friend cp operator + (cp a,cp b) {

        return cp(a.x+b.x,a.y+b.y);

    }

    friend cp operator - (cp a,cp b) {

        return cp(a.x-b.x,a.y-b.y);

    }

    friend cp operator * (cp a,cp b) {

        return cp(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);

    }

}A[mxn],B[mxn],C[mxn];

void FFT(cp *p,register int opt)

{

    for(register int i=0;i<lim;++i)

        if(i<r[i]) swap(p[i],p[r[i]]);

    for(register int mid=1;mid<lim;mid<<=1) {

        cp wn=cp(cos(PI/mid),opt*sin(PI/mid));

        for(register int len=mid<<1,j=0;j<lim;j+=len) {

            cp w=cp(1,0);

            for(register int k=0;k<mid;++k,w=w*wn) {

                cp x=p[j+k],y=w*p[j+mid+k];

                p[j+k]=x+y; p[j+mid+k]=x-y;

            }

        }

    }

}

int main()

{

    scanf("%d%d%s%s",&m,&n,s1,s2);

    for(register int i=0,j=m-1;i<j;++i,--j) swap(s1[i],s1[j]);

    for(register int i=0;i<m;++i) if(s1[i]!='*') a[i]=s1[i]-'a'+1;

    for(register int i=0;i<n;++i) if(s2[i]!='*') b[i]=s2[i]-'a'+1;

    while(lim<=n+m) ++l,lim<<=1;

    for(register int i=0;i<lim;++i)

        r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));

    for(register int i=0;i<=lim;++i)

        A[i]=cp(a[i]*a[i]*a[i],0),B[i]=cp(b[i],0);

    FFT(A,1); FFT(B,1);

    for(register int i=0;i<=lim;++i) C[i]=A[i]*B[i];

    FFT(C,-1);

    for(register int i=0;i<=lim;++i) s[i]+=(ll)(C[i].x/lim+0.5);

    for(register int i=0;i<=lim;++i)

        A[i]=cp(a[i],0),B[i]=cp(b[i]*b[i]*b[i],0);

    FFT(A,1); FFT(B,1);

    for(register int i=0;i<=lim;++i) C[i]=A[i]*B[i];

    FFT(C,-1);

    for(register int i=0;i<=lim;++i) s[i]+=(ll)(C[i].x/lim+0.5);

    for(register int i=0;i<=lim;++i)

        A[i]=cp(a[i]*a[i],0),B[i]=cp(b[i]*b[i],0);

    FFT(A,1); FFT(B,1);

    for(register int i=0;i<=lim;++i) C[i]=A[i]*B[i];

    FFT(C,-1);

    for(register int i=0;i<=lim;++i) s[i]-=2*(ll)(C[i].x/lim+0.5);

    for(register int i=m-1;i<n;++i) if(s[i]==0) q[++tot]=i-m+2;

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

    for(register int i=1;i<=tot;++i) printf("%d ",q[i]);

    return 0;

}