P4173 残缺的字符串（FFT）

Posted 2021-02-12 lizehon

[Luogu4173]

题解

(1.)定义匹配函数

(2.)定义完全匹配函数

(3.)快速计算每一位的完全匹配函数值

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define Debug(x) cout<<#x<<"="<<x<<endl
using namespace std;
typedef long long LL;
const int INF=1e9+7;
inline LL read(){
    register LL x=0,f=1;register char c=getchar();
    while(c<48||c>57){if(c=='-')f=-1;c=getchar();}
    while(c>=48&&c<=57)x=(x<<3)+(x<<1)+(c&15),c=getchar();
    return f*x;
}
inline char readc(){
    register char c=getchar();
    while(c==' '||c=='
'||c=='	') c=getchar();
    return c;
}

const int MAXN=2e6+5;
const double Pi=acos(-1);

struct cmpx{
    double x,y;
    inline cmpx(){}
    inline cmpx(double _x,double _y){x=_x,y=_y;}
    inline friend cmpx operator + (cmpx a,cmpx b){return cmpx(a.x+b.x,a.y+b.y);}
    inline friend cmpx operator - (cmpx a,cmpx b){return cmpx(a.x-b.x,a.y-b.y);}
    inline friend cmpx operator * (cmpx a,cmpx b){return cmpx(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
    inline friend cmpx operator * (cmpx a,double b){return cmpx(a.x*b,a.y*b);}
}A[MAXN],B[MAXN],C[MAXN];

char s[MAXN],t[MAXN];
int ans[MAXN],a[MAXN],b[MAXN];
int n,m;

namespace F_F_T{
    int rev[MAXN],limit,l;
    inline void init(int n){
        for(limit=1,l=0;limit<=n;limit<<=1) l++;
        for(int i=0;i<limit;i++)
            rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));
    }
    inline void FFT(cmpx *A,int type){
        for(int i=0;i<limit;i++)
            if(i<rev[i]) swap(A[i],A[rev[i]]);
        for(int len=1;len<limit;len<<=1){
            cmpx Wn=(cmpx){cos(Pi/len),type*sin(Pi/len)};
            for(int i=0;i<limit;i+=(len<<1)){
                cmpx w=(cmpx){1,0};
                for(int j=0;j<len;j++,w=w*Wn){
                    cmpx x=A[i+j],y=w*A[i+len+j];
                    A[i+j]=x+y;
                    A[i+len+j]=x-y;
                }
            }
        }
        if(type==-1){
            for(int i=0;i<limit;i++) A[i].x/=limit;
        }
    }
}using namespace F_F_T;

int main(){
    n=read(),m=read();
    init(n+m-2);
    for(int i=n-1;i>=0;i--){
        char c=readc();
        a[i]=(c!='*')?c-'a'+1:0;
    }
    for(int i=0;i<m;i++){
        char c=readc();
        b[i]=(c!='*')?c-'a'+1:0;
    }

    for(int i=0;i<n;i++) A[i]=cmpx(a[i]*a[i]*a[i],0);
    for(int i=0;i<m;i++) B[i]=cmpx(b[i],0);
    FFT(A,1);FFT(B,1);
    for(int i=0;i<limit;i++) C[i]=A[i]*B[i];

    for(int i=0;i<limit;i++) A[i]=B[i]=cmpx(0,0);
    for(int i=0;i<n;i++) A[i]=cmpx(a[i],0);
    for(int i=0;i<m;i++) B[i]=cmpx(b[i]*b[i]*b[i],0);
    FFT(A,1);FFT(B,1);
    for(int i=0;i<limit;i++) C[i]=C[i]+A[i]*B[i];

    for(int i=0;i<limit;i++) A[i]=B[i]=cmpx(0,0);
    for(int i=0;i<n;i++) A[i]=cmpx(a[i]*a[i],0);
    for(int i=0;i<m;i++) B[i]=cmpx(b[i]*b[i],0);
    FFT(A,1);FFT(B,1);
    for(int i=0;i<limit;i++) C[i]=C[i]-A[i]*B[i]*2.0;

    FFT(C,-1);

    for(int i=n-1;i<m;i++){
        if(fabs(C[i].x)<0.5) ans[++ans[0]]=i-n+2;
        //printf("%.3lf
",C[i].x);
    }
    printf("%d
",ans[0]);
    for(int i=1;i<=ans[0];i++)
        printf("%d ",ans[i]);
}