2019 Multi-University Training Contest 2 I.I Love Palindrome String（回文自动机+字符串hash）

Problem Description

You are given a string S=s1s2..s|S| containing only lowercase English letters. For each integer i∈[1,|S|] , please output how many substrings slsl+1...sr satisfy the following conditions:

 ∙ r−l+1 equals to i.

 ∙ The substring slsl+1...sr is a palindrome string.

 ∙ slsl+1...s⌊(l+r)/2⌋ is a palindrome string too.

|S| denotes the length of string S.

A palindrome string is a sequence of characters which reads the same backward as forward, such as madam or racecar or abba. 

 

Input

There are multiple test cases.

Each case starts with a line containing a string S(1≤|S|≤3×105) containing only lowercase English letters.

It is guaranteed that the sum of |S| in all test cases is no larger than 4×106.

 

Output

For each test case, output one line containing |S| integers. Any two adjacent integers are separated by a space.

 

Sample Input

abababa

 

Sample Output

7 0 0 0 3 0 0

 

题意:给你一个字符串 要你输出长度分别为1~len的good回文串的个数（good回文串定义为 当前是回文串 且他的一半也是回文串）

思路:我们首先用回文自动机把回文串都记录下来 然后字符串hash判断是否两半是否相等

#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
const int inf = 0x3f3f3f3f;
const int N = 4e5+7;
const ll mod = 998244353;
using namespace std;
ull hash1=233;
ull ha[N],pp[N];
ull getha(int l,int r){
    if(l==0) return ha[r];
    return ha[r]-ha[l-1]*pp[r-l+1];
}
struct Palindromic_Tree{
    int next[N][30]; //节点之间连边 
    int fail[N]; //适配指针 表示当前回文串的最长回文后缀 
    int len[N]; //当前回文串的长度 
    int cnt[N]; //回文串的个数 
    int id[N]; //回文串的右端点 
    int S[N];
    int last,n,p;
    int newnode(int l){//新建节点
        for(int i=0;i<26;i++) next[p][i]=0;//新建的节点为p，先消除它的子节点
        cnt[p]=0;
        len[p]=l;
        return p++;//勿打成++p，因为此节点为p，我们应返回p
    }
    void init(){
        last=n=p=0;
        newnode(0); newnode(-1);
        S[0]=-1; fail[0]=1;
    }
    int get_fail(int x){
        while(S[n-len[x]-1]!=S[n])    x=fail[x];
        return x;
    }
    void add(int c){
        c-='a';
        S[++n]=c;
        int po=get_fail(last);
        if(!next[po][c]){
            int now=newnode(len[po]+2);
            fail[now]=next[get_fail(fail[po])][c];
            next[po][c]=now;
        }
        last=next[po][c];
        cnt[last]++;
        id[last]=n;
    }
    void count(){
        for(int i=p-1;i>=0;i--)
            cnt[fail[i]]+=cnt[i];
    }
}pat;
ll ans[N];
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    pp[0]=1;
    for(int i=1;i<N;i++) {
        pp[i]=hash1*pp[i-1];
    }
    string s;
    while(cin>>s){
        memset(ans,0,sizeof(ans));
        pat.init();
        int len=s.length();
        for(int i=0;i<len;i++){
            pat.add(s[i]);
        }
        ha[0]=s[0];
        for(int i=1;i<len;i++)
            ha[i]=ha[i-1]*hash1+s[i];
        pat.count();
        for(int i=2;i<pat.p;i++){
            int l=pat.id[i]-pat.len[i];
            int r=pat.id[i]-1;
            int mid=(l+r)>>1;
            if((r-l+1)&1){
                if(getha(l,mid)==getha(mid,r)){
                    ans[pat.len[i]]+=pat.cnt[i];
                }    
            }else{
                if(getha(l,mid)==getha(mid+1,r)){
                    ans[pat.len[i]]+=pat.cnt[i];
                }
            }
        }
        for(int i=1;i<=len;i++){
            if(i==1) cout<<ans[i];
            else cout<<" "<<ans[i];
        }
        cout<<endl;
    }
    return 0;
}