BZOJ4503: 两个串

Posted 2020-08-30

传送门

FFT竟然还能解决字符串问题..

但是这个数据好像太水了，KMP瞎搞搞就能过。

考虑一下，如果不考虑问号，那么一个点匹配肯定存在的是$\sum (s1_i-s2_j)^2=0$，那么考虑存在？的情况。很显然，改成这样的即可：

$\sum (s1_i-s2_j)^2 s2_j$

化开即为$\sum s1_i^2s2_j+s2_j^3-2 \times s1_is2_j^2$

可以把$s2$reverse，然后$[M-1,N-1]$即为答案。

//BZOJ4503
//by Cydiater
//2017.2.11
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <iomanip>
#include <algorithm>
#include <bitset>
#include <set>
#include <vector>
#include <complex>
using namespace std;  
#define ll 		long long
#define up(i,j,n)	for(int i=j;i<=n;i++)
#define down(i,j,n)	for(int i=j;i>=n;i--)
#define cmax(a,b)	a=max(a,b)
#define cmin(a,b)	a=min(a,b)
#define db 		double
#define Plur		complex<double>
const int MAXN=5e5+5;
const int oo=0x3f3f3f3f;
const db PI=acos(-1);
char s1[MAXN],s2[MAXN];
Plur A[MAXN],B[MAXN],A2[MAXN],B2[MAXN],P1[MAXN],P2[MAXN],P3[MAXN];
int N,M,NM,ans=0;
namespace FFT{
	Plur omega[MAXN],Invomega[MAXN];
	void Prepare(int LIM){
		up(i,0,LIM-1){
			omega[i]=Plur(cos(2*PI*i/LIM),sin(2*PI*i/LIM));
			Invomega[i]=conj(omega[i]);
		}
	}
	void Transform(int N,Plur *A,Plur *w){
		int pos=0;
		up(i,0,N-1){
			if(pos<i)swap(A[i],A[pos]);
			for(int tmp=N>>1;(pos^=tmp)<tmp;tmp>>=1);
		}
		for(int l=2;l<=N;l<<=1){
			int delta=l>>1;
			for(int j=0;j<N;j+=l)up(k,0,delta-1){
				Plur tmp=A[j+k+delta]*w[N/l*k];
				A[j+delta+k]=A[j+k]-tmp;
				A[j+k]+=tmp;
			}
		}
	}
	void DFT(int N,Plur *A){
		Transform(N,A,omega);
	}
	void IDFT(int N,Plur *A){
		Transform(N,A,Invomega);
		up(i,0,N-1)A[i]/=N;
	}
}
namespace solution{
	void Prepare(){
		scanf("%s",s1);scanf("%s",s2);
		N=strlen(s1);M=strlen(s2);
		reverse(s2,s2+M);
		NM=N+M-1;
		up(i,0,M-1)if(s2[i]==‘?‘)s2[i]=‘a‘-1;
		int ind=0;
		while((1<<ind)<NM)ind++;
		NM=(1<<ind);
		up(i,0,N-1)A[i].real(s1[i]-‘a‘+1);
		up(i,0,N-1)A2[i].real((s1[i]-‘a‘+1)*(s1[i]-‘a‘+1));
		up(i,0,M-1)B[i].real(s2[i]-‘a‘+1);
		up(i,0,M-1)B2[i].real((s2[i]-‘a‘+1)*(s2[i]-‘a‘+1));
	}
	int col(int pos){return (int)(P1[pos].real()+P2[pos].real()-P3[pos].real()-P3[pos].real()+0.5);}
	void Solve(){
		FFT::Prepare(NM);
		FFT::DFT(NM,A);FFT::DFT(NM,A2);
		FFT::DFT(NM,B);FFT::DFT(NM,B2);
		up(i,0,NM-1)P1[i]=A2[i]*B[i];
		up(i,0,NM-1)P3[i]=A[i]*B2[i];
		FFT::IDFT(NM,P1);
		FFT::IDFT(NM,P3);
		int tmp=0;
		up(i,0,M-1)tmp+=(s2[i]-‘a‘+1)*(s2[i]-‘a‘+1)*(s2[i]-‘a‘+1);
		up(i,0,NM-1)P2[i].real(tmp);
		up(i,M-1,N-1)if(col(i)==0)ans++;
		cout<<ans<<endl;
		up(i,M-1,N-1)if(col(i)==0){
			printf("%d\n",i-(M-1));
		}
	}
}
int main(){
	//freopen("input.in","r",stdin);
	using namespace solution;
	Prepare();
	Solve();
	return 0;
}