[hdu6599]I Love Palindrome String

Time Limit: 4000/2000 MS (Java/Others)
Memory Limit: 131072/131072 K (Java/Others)
分数：回文自动机初见，2500
Problem Description

You are given a string S=s1s2..s∣S∣S=s_1s_2..s_{|S|}S=s1s2..s∣S∣ containing only lowercase English letters. For each integer i∈[1,∣S∣]i∈[1,|S|]i∈[1,∣S∣], please output how many substrings slsl+1...srs_ls_{l+1}...s_rslsl+1...sr satisfy the following conditions:
∙ r−l+1r−l+1r−l+1 equals to iii.
∙ The substring slsl+1...srs_ls_{l+1}...s_rslsl+1...sr is a palindrome string.
∙ slsl+1...s⌊(l+r)/2⌋s_ls_{l+1}...s_{⌊(l+r)/2⌋}slsl+1...s⌊(l+r)/2⌋ is a palindrome string too.

∣S∣|S|∣S∣ denotes the length of string SSS.

A palindrome string is a sequence of characters which reads the same backward as forward, such as madammadammadam or racecarracecarracecar or abbaabbaabba.

Input
There are multiple test cases.
Each case starts with a line containing a string S(1≤∣S∣≤3×105)S(1≤|S|≤3×10^5)S(1≤∣S∣≤3×105) containing only lowercase English letters.
It is guaranteed that the sum of ∣S∣|S|∣S∣ in all test cases is no larger than 4×1064×10^64×106.

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

Sample Input

abababa

Sample Output

7 0 0 0 3 0 0

Source

2019 Multi-University Training Contest 2

题意：
给定一个字符串，问字符串中长度为i(1<=i<=n)i(1<=i<=n)i(1<=i<=n)的符合下列条件的子串的个数
条件：
1.这个串是个回文串
2.这个串的前半部分（如果长度为奇数则要算上最中间的字符）也是回文串

题解：
枚举本质不同的回文串：回文自动机
判断一个字符串是不是回文串：哈希
枚举每种本质不同的回文串，我们可以用哈希O(1)判断他的前半部分是不是也是回文串。
然后用PAM中的每个节点的size计数即可。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll MOD=19991227;
const int MAXN=300004;
const int D=27;
char s[MAXN];
ll bin[MAXN],Hash[2][MAXN];
int n;
void workHash(){Hash[0][0]=0;for(int i=1;i<=n;i++){Hash[0][i]=(Hash[0][i-1]*D+(s[i]-'a'+1))%MOD;}Hash[1][n+1]=0;for(int i=n;i>=1;i--){Hash[1][i]=(Hash[1][i+1]*D+(s[i]-'a'+1))%MOD;}
}
ll get(int l,int r,int x){if(!x){return (Hash[x][r]-((Hash[x][l-1]*bin[r-l+1])%MOD)+MOD)%MOD;}else{return (Hash[x][l]-((Hash[x][r+1]*bin[r-l+1])%MOD)+MOD)%MOD;}
}
bool check(int l,int r){return (get(l,(l+r)>>1,0)==get(l,(l+r)>>1,1));
}
struct PAM{int last,cnt;int len[MAXN],ch[MAXN][D],sz[MAXN],fa[MAXN],exi[MAXN];int ans[MAXN];int create(int Len,int Fa){len[cnt]=Len;fa[cnt]=Fa;return cnt++;}void CLEAR(){for(int i=0;i<cnt;i++){fa[i]=0;len[i]=0;sz[i]=0;memset(ch[i],0,sizeof(ch[i]));}cnt=0;last=0;create(0,1);create(-1,0);}int getfail(int p,int n){for(;s[n-len[p]-1]!=s[n];p=fa[p]);return p;}int add(int c,int pos){int p=getfail(last,pos);if(!ch[p][c]){ch[p][c]=create(len[p]+2,ch[getfail(fa[p],pos)][c]);exi[ch[p][c]]=check(pos-len[ch[p][c]]+1,pos);}last=ch[p][c];sz[last]++;return last;}void getans(){for(int i=cnt-1;i>=2;i--){sz[fa[i]]+=sz[i];}for(int i=cnt-1;i>=2;i--){ans[len[i]]+=sz[i]*exi[i];}for(int i=1;i<=n;i++){printf("%d",ans[i]);if(i==n)puts("");else printf(" ");}}
}pam;
int w33ha(){n=strlen(s+1);workHash();pam.CLEAR();for(int i=1;i<=n;i++)pam.ans[i]=0;for(int i=1;i<=n;i++)pam.add(s[i]-'a',i);pam.getans();return 0;
}
int main(){bin[0]=1;for(int i=1;i<=MAXN;i++)bin[i]=(bin[i-1]*D)%MOD;while(scanf("%s",s+1)!=EOF)w33ha();return 0;
}

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