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 }