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URAL1960 Palindromes and Super Abilities

2017-02-17 18:29 393 查看

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……

/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
const int mxn=102100;
int read(){
int x=0,f=1;char ch=getchar();
while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
char s[mxn];
struct Pam{
int t[mxn][27];
int l[mxn],sz[mxn],fa[mxn];
int res[mxn];
int S,cnt,last;
void init(){S=cnt=last=fa[0]=fa[1]=1;l[1]=-1;return;}
void add(int c,int n){
int p=last;
while(s[n-l-1]!=s
)p=fa[p];
if(!t[p][c]){
int np=++cnt;l[np]=l[p]+2;
int k=fa[p];
while(s[n-l[k]-1]!=s
)k=fa[k];
fa[np]=t[k][c];//fail指针
t[p][c]=np;
}
last=t[p][c];
sz[last]++;//统计出现次数
}
void solve(){
printf("%d ",cnt-1);
/*
memset(res,0,sizeof res);
int ans=0;
for(int i=cnt;i;i--){
res[i]+=sz[i];
res[fa[i]]+=res[i];
ans+=res[i];
}
printf("%d ",ans);*/
return;
}
}hw;
int main(){
int i,j;
scanf("%s",s+1);
int len=strlen(s+1);
hw.init();
for(i=1;i<=len;i++){
hw.add(s[i]-'a',i);
hw.solve();
}
return 0;
}

[p]

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