CF954I Yet Another String Matching Problem （ FFT ）

Description

is given two strings S and T
the distance between the two strings A and B of equal length defined as follows:
each operator may select two characters c1 and c2, c1 replace all strings of two c2, so that a and B become equal to the minimum number of operations.
Each seek length S is | T | sub-string and the distance between the T.
n <= 125000, S and T to ensure that contain only lowercase letters a ~ f

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Violence, then string with disjoint-set o (n), then sweep over each sub complexity of s × t is

Consider how to optimize

The title character set of six, so you can consider in each piece, whether at the corresponding position matched (this may do the FFT) between two characters enumeration

Ruoguo matching words, look in the current block, whether the two characters FA is a

 

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 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<set>
 8 #include<map>
 9 #include<vector>
10 #include<queue>
11 using namespace std;
12 #define ll long long
13 #define RG register
14 #define MAX 333333
15 const double Pi=acos(-1);
16 struct Complex{double a,b;}A[MAX],B[MAX],W[MAX];
17 Complex operator+(Complex a,Complex b){return (Complex){a.a+b.a,a.b+b.b};}
18 Complex operator-(Complex a,Complex b){return (Complex){a.a-b.a,a.b-b.b};}
19 Complex operator*(Complex a,Complex b){return (Complex){a.a*b.a-a.b*b.b,a.a*b.b+a.b*b.a};}
20 int r[MAX],N,n,m,l,eql[MAX][6][6];
21 char a[MAX],b[MAX];
22 void FFT(Complex *P,int opt)
23 {
24     for(int i=1;i<N;++i)if(i<r[i])swap(P[i],P[r[i]]);
25     for(int i=1;i<N;i<<=1)
26         for(int p=i<<1,j=0;j<N;j+=p)
27             for(int k=0;k<i;++k)
28             {
29                 Complex w=(Complex){W[N/i*k].a,W[N/i*k].b*opt};
30                 Complex X=P[j+k],Y=w*P[i+j+k];
31                 P[j+k]=X+Y;P[i+j+k]=X-Y;
32             }
33 }
34 int f[6];
35 int getf(int x){return x==f[x]?x:f[x]=getf(f[x]);}
36 int main()
37 {
38     scanf("%s",a);scanf("%s",b);
39     n=strlen(a),m=strlen(b);
40     for(N=1;N<=(n+m);N<<=1)++l;
41     for(int i=0;i<N;++i)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
42     for(int i=1;i<N;i<<=1)
43         for(int k=0;k<i;++k)W[N/i*k]=(Complex){cos(k*Pi/i),sin(k*Pi/i)};
44     for(int i=0;i<6;++i)
45         for(int j=0;j<6;++j)
46         {
47             for(int k=0;k<N;++k)A[k].a=A[k].b=B[k].a=B[k].b=0;
48             for(int k=0;k<n;++k)A[k].a=(a[k]==i+97);
49             for(int k=0;k<m;++k)B[k].a=(b[m-k-1]==j+97);
50             FFT(A,1);FFT(B,1);
51             for(int k=0;k<N;++k)A[k]=A[k]*B[k];
52             FFT(A,-1);
53             for(int k=0;k<N;++k)eql[k][i][j]=(int)(A[k].a/N+0.5);
54         }
55     for(int i=m-1;i<n;++i)
56     {
57         for(int j=0;j<6;++j)f[j]=j;
58 
59         int ans=0;
60 
61         for(int j=0;j<6;++j)
62             for(int k=0;k<6;++k)
63                 if(eql[i][j][k])
64                    {
65                     if(getf(j) != getf(k))
66                     {
67                         ans ++ ;
68                         f[getf(j)] = getf(k);
69                     }
70                    }
71 
72         printf("%d ",ans);
73     }
74     puts("");
75     return 0;
76 }