Up Sky,Mr.Zhu

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 46    Accepted Submission(s): 14

Problem Description Given a string S[0,…n-1],and the length of each palindrome substring in S is less than 20.

Define special string of a palindrome string STR[0,…,n-1] is STR[floor(n/2),…,n-1]

Given L,R and a string T, Query the number of palindrome strings, whose the prefix of special string is T, in S[L,…,R] 

ps:floor means the function in C++ language

(No relation between statement and title)  

Input There are severval cases,please peocess until EOF.

The first line contains string S.(1<=|S|<=1e5)

The second line is a number q means the number of querys.(1<=q<=1e5)

The next q lines,each line contains two numbers L,R and a string T.(1<=L<=R<=|S|,1<=|T|<=10)

All strings only contain ‘a’,’b’,’c’,’d’ and ‘e’.  

Output For each query,you should output the number of palindrome strings which satisfied previous condition in a line.  

Sample Input

bceaeedde 5 5 8 e 3 5 e 1 2 a 5 9 d 5 9 de  

Sample Output

3 2 0 4 1  

Author UESTC  

Source
2016 Multi-University Training Contest 6

题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=5801

题目大意：
定义一个回文串的特征串为其从中心到结尾的部分，给出一个字符串S，q次询问，每次询问以一个字符串t为特征串前缀落在区间L到R之间的回文串数量，S中回文串长度不会超过20。

解题思路：
先Manacher预处理，然后我们用Trie树来维护以每个位置为起始的特征串，注意到特征串的长度不会超过10，所以我们可以每次按一个长度建立可持久化Trie树，每次对所有询问的区间在线查询，更新答案，做10次即可。
为了不去考虑角码的运算，我将奇偶串分开讨论了，代码就显得有些冗余。最后要吐槽hdu谜一样的java内存机制，开始估计数组最大要开到10^6，结果MLE了，索性开到10^5枚举内存，居然神奇的AC了。

AC代码：

import java.io.*;
import java.util.*;

public class Main {
    static StreamTokenizer in=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    static int nextInt() throws IOException    
    {  
        in.nextToken();    
        return (int)in.nval;     
    }
    static String next() throws IOException    
    {  
        in.nextToken();  
        return in.sval;
    }
    static PrintWriter out=new PrintWriter(new OutputStreamWriter(System.out));
    static int n,m,len,sz,a;
    static char[] s;
    static char[] ss=new char[200005];
    static int[] pp=new int[200005];
    static int[] aa=new int[100005];
    static int[] bb=new int[100005];
    static int[] cc=new int[100005];
    static int[] rt=new int[100005];
    static int[] ans=new int[100005];
    static int[] cnt=new int[100005];
    static char[][] tt=new char[100005][];
    static int[][] ch=new int[100005][5];
    static void Manacher()
    {
        Arrays.fill(ss,'#');
        for(int i=0;i<n;i++)
            ss[i*2+2]=s[i];
        len=n*2+1;ss[0]='$';
        int max=0,id=0;
        for(int i=1;i<=len;i++)
        {
            if(max>i)
                pp[i]=Math.min(pp[id*2-i],max-i);
            else pp[i]=1;
            while(ss[i+pp[i]]==ss[i-pp[i]])
                pp[i]++;
            if(i+pp[i]>max) { max=i+pp[i];id=i;}
        }
    }
    static void insert(int a,int b,int l,int r)
    {
        int u=a,v=b,p;
        for(int i=l;i<=r;i++)
        {
            p=ss[i]-'a';
            for(int k=0;k<5;k++)
                ch[u][k]=ch[v][k];
            cnt[u]=cnt[v]+1;
            ch[u][p]=++sz;
            u=ch[u][p];v=ch[v][p];
        }
        cnt[u]=cnt[v]+1;
    }
    static int query(int a,int b,int k)
    {
        int u=a,v=b,p;
        for(int i=0;i<tt[k].length;i++)
        {
            p=tt[k][i]-'a';
            u=ch[u][p];v=ch[v][p];
        }
        return cnt[v]-cnt[u];
    }

    public static void main(String[] args) throws IOException {
        //Scanner in=new Scanner(System.in);
        while(in.nextToken()!=StreamTokenizer.TT_EOF)
        {
            s=in.sval.toCharArray();
            n=s.length;
            Manacher();
            for(int i=n;i>0;i--) ss[i]=s[i-1];
            m=nextInt();
            for(int i=1;i<=m;i++)
            {
                aa[i]=nextInt();bb[i]=nextInt();
                tt[i]=next().toCharArray();
                cc[i]=tt[i].length;
            }
            for(int i=1;i<=m;i++) ans[i]=0;
            for(int k=1;k<=10;k++)
            {
                sz=0;rt[0]=++sz;
                for(int i=1;i<=n;i++)
                {
                    a=pp[i*2]/2;
                    if(a>=k)
                    {
                        rt[i]=++sz;
                        insert(rt[i],rt[i-1],i,i+k-1);
                    }
                    else rt[i]=rt[i-1];
                }
                for(int i=1;i<=m;i++)
                {
                    if(cc[i]>k||k*2-1>bb[i]-aa[i]+1) continue;
                    ans[i]+=query(rt[aa[i]+k-2],rt[bb[i]-k+1],i);
                }
                sz=0;rt[0]=++sz;
                for(int i=1;i<=n;i++)
                {
                    a=pp[i*2-1]/2;
                    if(a>=k)
                    {
                        rt[i]=++sz;
                        insert(rt[i],rt[i-1],i,i+k-1);
                    }
                    else rt[i]=rt[i-1];
                }
                for(int i=1;i<=m;i++)
                {
                    if(cc[i]>k||k*2>bb[i]-aa[i]+1) continue;
                    ans[i]+=query(rt[aa[i]+k-1],rt[bb[i]-k+1],i);
                }
            }
            for(int i=1;i<=m;i++)
                out.println(ans[i]);
        }
        out.flush();
    }
}