P4173 残缺的字符串 fft

Posted 2020-12-25 acjiumeng

题意:给你两个字符串,问你第一个在第二个中出现过多少次,并输出位置,匹配时是模糊匹配*可和任意一个字符匹配
题解:fft加速字符串匹配;
假设上面的串是s,s长度为m,下面的串是p,p长度为n,先考虑没有*的情况那么(sum_{j=1}^m(s_{i+j}-p_j)^2=0)就表示能够从i开始匹配,现在考虑有*的情况,我们只需要让有*的和任意字符匹配即可,那么把公式变成(sum_{j=1}^m(s_{i+j}-p_j)^2*s_{i+j}*p_j)=0),但是fft正向匹配太慢了,我们把方向变一下,(sum_{j=1}^m(s_{i+j}-p_{n-j})^2*s_{i+j}*p_{n-j}=0),把式子分解一下得(sum_{j=1}^m(s_{i+j}^3*p_{n-j}+s_{i+j}*p_{n-j}^3-2*s_{i+j}^2*p_{n-j}^2)=0),这样把这6个多项式跑一边fft,就得到了答案,

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=300000+10,maxn=50000+10,inf=0x3f3f3f3f;

struct cd{
    db x,y;
    cd(db _x=0.0,db _y=0.0):x(_x),y(_y){}
    cd operator +(const cd &b)const{
        return cd(x+b.x,y+b.y);
    }
    cd operator -(const cd &b)const{
        return cd(x-b.x,y-b.y);
    }
    cd operator *(const cd &b)const{
        return cd(x*b.x - y*b.y,x*b.y + y*b.x);
    }
    cd operator /(const db &b)const{
        return cd(x/b,y/b);
    }
}a[N<<3],b[N<<3],c[N<<3],d[N<<3],e[N<<3],f[N<<3];
int rev[N<<3];
void getrev(int bit)
{
    for(int i=0;i<(1<<bit);i++)
        rev[i]=(rev[i>>1]>>1) | ((i&1)<<(bit-1));
}
void fft(cd *a,int n,int dft)
{
    for(int i=0;i<n;i++)
        if(i<rev[i])
            swap(a[i],a[rev[i]]);
    for(int step=1;step<n;step<<=1)
    {
        cd wn(cos(dft*pi/step),sin(dft*pi/step));
        for(int j=0;j<n;j+=step<<1)
        {
            cd wnk(1,0);
            for(int k=j;k<j+step;k++)
            {
                cd x=a[k];
                cd y=wnk*a[k+step];
                a[k]=x+y;a[k+step]=x-y;
                wnk=wnk*wn;
            }
        }
    }
    if(dft==-1)for(int i=0;i<n;i++)a[i]=a[i]/n;
}
char s[N],p[N];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    scanf("%s%s",s+1,p+1);
    int sz=0;
    while((1<<sz)<m)sz++;
    sz++,getrev(sz);
    for(int i=0;i<=(1<<sz);i++)
        a[i]=b[i]=c[i]=d[i]=e[i]=f[i]=0;
    for(int i=1,te;i<=n;i++)
    {
        if(s[i]=='*')te=0;
        else te=s[i]-'a'+1;
        a[m-i]=te*te*te,b[m-i]=-2*te*te,c[m-i]=te;
    }
    for(int i=1,te;i<=m;i++)
    {
        if(p[i]=='*')te=0;
        else te=p[i]-'a'+1;
        d[i]=te*te*te,e[i]=te*te,f[i]=te;
    }
    fft(a,(1<<sz),1),fft(b,(1<<sz),1),fft(c,(1<<sz),1);
    fft(d,(1<<sz),1),fft(e,(1<<sz),1),fft(f,(1<<sz),1);
    for(int i=0;i<=(1<<sz);i++)
        d[i]=a[i]*f[i]+b[i]*e[i]+c[i]*d[i];
    fft(d,(1<<sz),-1);
//    for(int i=0;i<=m-n;i++)
//        printf("%d ",(int)((d[m+i].x+0.5)/(1<<sz)));
//    puts("");
    vi ans;
    for(int i=0;i<=m-n;i++)
        if((int)(d[m+i].x+0.5/(1<<sz))==0)
            ans.pb(i+1);
    printf("%d
",ans.size());
    for(int i=0;i<ans.size();i++)
        printf("%d ",ans[i]);
    puts("");
    return 0;
}
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