BZOJ3160:万径人踪灭(FFT,Manacher)

Solution

$ans=$回文子序列$-$回文子串的数目。

后者可以用$manacher$直接求。

前者设$f[i]$表示以$i$为中心的对称的字母对数。

那么回文子序列的数量也就是$\sum_{i=0}^{n-1}2^{f[i]-1}$

构造两个数组$a[i],b[i]$。若第$i$位为$a$,那么$a[i]=1$,否则$b[i]=1$。

可以发现$a$数组自身卷积就是$a$字母对$f$数组的贡献,$b$数组同理。

卷下$a$,卷下$b$,对应位置求和,就是$f$数组。

因为在卷积中每对对称字符被算了两次,而自己和自己关于自己对称只算了一次,所以要把答案除2向上取整。

Code

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<cmath>
 5 #define N (400009)
 6 #define LL long long
 7 #define MOD (1000000007)
 8 using namespace std;
 9 
10 int n,fn,l,tot,r[N],len[N],p[N];
11 LL Re,fun;
12 char s[N],st[N];
13 double pi=acos(-1.0);
14 struct complex
15 {
16     double x,y;
17     complex (double xx=0,double yy=0)
18     {
19         x=xx; y=yy;
20     }
21 }a[N],b[N];
22 
23 complex operator + (complex a,complex b) {return complex(a.x+b.x,a.y+b.y);}
24 complex operator - (complex a,complex b) {return complex(a.x-b.x,a.y-b.y);}
25 complex operator * (complex a,complex b) {return complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
26 complex operator / (complex a,double b) {return complex(a.x/b,a.y/b);}
27 
28 void FFT(int n,complex *a,int opt)
29 {
30     for (int i=0; i<n; ++i)
31         if (i<r[i]) swap(a[i],a[r[i]]);
32     for (int k=1; k<n; k<<=1)
33     {
34         complex wn=complex(cos(pi/k),opt*sin(pi/k));
35         for (int i=0; i<n; i+=k<<1)
36         {
37             complex w=complex(1,0);
38             for (int j=0; j<k; ++j,w=w*wn)
39             {
40                 complex x=a[i+j], y=w*a[i+j+k];
41                 a[i+j]=x+y; a[i+j+k]=x-y;
42             }
43         }
44     }
45     if (opt==-1) for (int i=0; i<n; ++i) a[i]=a[i]/n;
46 }
47 
48 void Manacher()
49 {
50     s[++tot]='('; s[++tot]='#';
51     for (int i=0; i<n; ++i)
52         s[++tot]=st[i], s[++tot]='#';
53     s[++tot]=')';
54     int maxn=0,mid=0,x;
55     for (int i=1; i<=tot; ++i)
56     {
57         if (i>maxn) x=1;
58         else x=min(maxn-i+1,len[mid*2-i]);
59         while (s[i+x]==s[i-x]) x++;
60         len[i]=x;
61         if (i+x-1>maxn) maxn=i+x-1, mid=i;
62         fun=(fun+len[i]/2)%MOD;
63     }
64 }
65 
66 int main()
67 {
68     p[0]=1;
69     for (int i=1; i<=100000; ++i)
70         p[i]=p[i-1]*2%MOD;
71     scanf("%s",st);    n=strlen(st);
72     Manacher();
73 
74     fn=1;
75     while (fn<=n+n) fn<<=1, l++;
76     for (int i=0; i<fn; ++i)
77         r[i]=(r[i>>1]>>1) | ((i&1)<<(l-1));
78     for (int i=0; i<n; ++i)
79         if (st[i]=='a') a[i].x=1;
80         else b[i].x=1;
81     FFT(fn,a,1); FFT(fn,b,1);
82     for (int i=0; i<fn; ++i)
83         a[i]=a[i]*a[i], b[i]=b[i]*b[i];
84     FFT(fn,a,-1); FFT(fn,b,-1);
85     for (int i=0; i<fn; ++i)
86     {
87         int x=(a[i].x+b[i].x+0.5);
88         x=(x+1)>>1;
89         Re=(Re+p[x]-1)%MOD;
90     }
91     printf("%lld\n",(Re-fun+MOD)%MOD);
92 }
posted @ 2018-12-09 15:50  Refun  阅读(177)  评论(0编辑  收藏  举报