字符串 回文自动机
After solving seven problems on Timus Online Judge with a word “palindrome” in the problem name, Misha has got an unusual ability. Now, when he reads a word, he can mentally count the number of unique nonempty substrings of this word that are palindromes.
Dima wants to test Misha’s new ability. He adds letters s 1, ..., s n to a word, letter by letter, and after every letter asks Misha, how many different nonempty palindromes current word contains as substrings. Which n numbers will Misha say, if he will never be wrong?
Input
The only line of input contains the string s 1... s n, where s i are small English letters (1 ≤ n ≤ 10 5).
Output
Output n numbers separated by whitespaces, i-th of these numbers must be the number of different nonempty substrings of prefix s 1... s i that are palindromes.
Example
input | output |
aba | 1 2 3 |
按顺序每次添加一个字符，求当前本质不同的回文串的数量
回文自动机
刚开始没意识到“本质不同”，加了个统计出现次数……
1 /*by SilverN*/
2 #include<algorithm>
3 #include<iostream>
4 #include<cstring>
5 #include<cstdio>
6 #include<cmath>
7 #include<vector>
8 using namespace std;
9 const int mxn=102100;
10 int read(){
11 int x=0,f=1;char ch=getchar();
12 while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
13 while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
14 return x*f;
15 }
16 char s[mxn];
17 struct Pam{
18 int t[mxn][27];
19 int l[mxn],sz[mxn],fa[mxn];
20 int res[mxn];
21 int S,cnt,last;
22 void init(){S=cnt=last=fa[0]=fa[1]=1;l[1]=-1;return;}
23 void add(int c,int n){
24 int p=last;
25 while(s[n-l[p]-1]!=s[n])p=fa[p];
26 if(!t[p][c]){
27 int np=++cnt;l[np]=l[p]+2;
28 int k=fa[p];
29 while(s[n-l[k]-1]!=s[n])k=fa[k];
30 fa[np]=t[k][c];//fail指针
31 t[p][c]=np;
32 }
33 last=t[p][c];
34 sz[last]++;//统计出现次数
35 }
36 void solve(){
37 printf("%d ",cnt-1);
38 /*
39 memset(res,0,sizeof res);
40 int ans=0;
41 for(int i=cnt;i;i--){
42 res[i]+=sz[i];
43 res[fa[i]]+=res[i];
44 ans+=res[i];
45 }
46 printf("%d ",ans);*/
47 return;
48 }
49 }hw;
50 int main(){
51 int i,j;
52 scanf("%s",s+1);
53 int len=strlen(s+1);
54 hw.init();
55 for(i=1;i<=len;i++){
56 hw.add(s[i]-'a',i);
57 hw.solve();
58 }
59 return 0;
60 }
