[BZOJ4503]两个串（FFT）

题目描述

传送门

题解

受到上一道题的影响，这题想了很久… 因为这题的思路和上题真的是相似又互异的 令 F(i) 表示把小串结尾怼到大串的第i个字符上是否能匹配，0表示能匹配，非0表示不能匹配 对于大串， f(i)=s(i) ，对于小串， g(i)=(s(i)=′?′)?0:s(i) ，也就是说，将通配符的权赋为0，其余的直接赋为自己的权值 那么计算将小串倒置了之后 F(i) 就可以写成卷积的形式，为避免正负数相消，我们将差取平方 F(i)=∑j=0t−1g(j)[f(i−j)−g(j)]2=∑j=0t−1f(i−j)2g(j)−2f(i−j)g(j)2+g(j)3 就相当于是算两个卷积然后对于每一个 F(i) 加一个常数 FFT

代码

#include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<cmath> using namespace std; #define N 300005 const double pi=acos(-1.0); int s,t,L,R[N],F[N],ans[N]; char S[N],T[N]; double f[N],g[N],z; 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);} void FFT(complex a[N],int 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 n,int m,int opt) { L=0;memset(R,0,sizeof(R)); m+=n; 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)); FFT(a,n,1);FFT(b,n,1); for (int i=0;i<=n;++i) a[i]=a[i]*b[i]; FFT(a,n,-1); for (int i=0;i<s;++i) F[i]+=opt*(int)(a[i].x/n+0.5); } int main() { scanf("%s",S);s=strlen(S); scanf("%s",T);t=strlen(T); for (int i=0;i<t/2;++i) swap(T[i],T[t-i-1]); for (int i=0;i<s;++i) f[i]=(double)(S[i]-'a'+1); for (int i=0;i<t;++i) g[i]=(T[i]=='?')?0.0:(double)(T[i]-'a'+1); memset(a,0,sizeof(a));memset(b,0,sizeof(b)); for (int i=0;i<s;++i) a[i]=complex(f[i]*f[i],0); for (int i=0;i<t;++i) b[i]=complex(g[i],0); calc(s-1,t-1,1); memset(a,0,sizeof(a));memset(b,0,sizeof(b)); for (int i=0;i<s;++i) a[i]=complex(2.0*f[i],0); for (int i=0;i<t;++i) b[i]=complex(g[i]*g[i],0); calc(s-1,t-1,-1); for (int i=0;i<t;++i) z+=g[i]*g[i]*g[i]; for (int i=t-1;i<s;++i) { F[i]+=(int)z; if (!F[i]) ans[++ans[0]]=i-t+1; } printf("%d\n",ans[0]); for (int i=1;i<=ans[0];++i) printf("%d\n",ans[i]); }

总结

思路要灵活