CA Loves Palindromic

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Problem Description

CA loves strings, especially loves the palindrome strings.

One day he gets a string, he wants to know how many palindromic substrings in the substring S[l,r].

Attantion, each same palindromic substring can only be counted once.

Input

First line contains T denoting
the number of testcases.

T testcases
follow. For each testcase:

First line contains a string S.
We ensure that it is contains only with lower case letters.

Second line contains a interger Q,
denoting the number of queries.

Then Q lines
follow, In each line there are two intergers l,r,
denoting the substring which is queried.

1≤T≤10, 1≤length≤1000, 1≤Q≤100000, 1≤l≤r≤length

Output

For each testcase, output the answer in Q lines.

Sample Input

1
abba
2
1 2
1 3

Sample Output

2
3

求区间内的本质不同的回文串的个数

字符串的长度是1000

我们可以利用回文树，求出每个区间内不同回文串的个数

枚举区间
#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdlib.h>
#include <math.h>
#include <stdio.h>

using namespace std;
typedef long long int LL;
const int MAX=100000;
const int maxn=1000;
char str[maxn+5];
int sum[maxn+5][maxn+5];
struct Tree
{
    int next[MAX+5][26];
    int num[MAX+5];
    int cnt[MAX+5];
    int fail[MAX+5];
    int len[MAX+5];
    int s[MAX+5];
    int p;
    int last;
    int n;
    int new_node(int x)
    {
        memset(next[p],0,sizeof(next[p]));
        cnt[p]=0;
        num[p]=0;
        len[p]=x;
        return p++;
    }
    void init()
    {
        p=0;
        new_node(0);
        new_node(-1);
        last=0;
        n=0;
        s[0]=-1;
        fail[0]=1;
    }
    int get_fail(int x)
    {
        while(s[n-len[x]-1]!=s[n])
            x=fail[x];
        return x;
    }
    int add(int x)
    {
        x-='a';
        s[++n]=x;
        int cur=get_fail(last);
        if(!(last=next[cur][x]))
        {
            int now=new_node(len[cur]+2);
            fail[now]=next[get_fail(fail[cur])][x];
            next[cur][x]=now;
            num[now]=num[fail[now]]+1;
            last=now;
            return 1;
        }
        cnt[last]++;
        return 0;
    }
    void count()
    {
        for(int i=p-1;i>=0;p++)
            cnt[fail[i]]+=cnt[i];
    }
}tree;
int q;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {

        scanf("%s",str+1);

        int len=strlen(str+1);
        for(int i=1;i<=len;i++)
        {
            tree.init();
            for(int j=i;j<=len;j++)
            {
               tree.add(str[j]);
               sum[i][j]=tree.p-2;
            }
        }
        scanf("%d",&q);
        int l,r;
        for(int i=1;i<=q;i++)
        {
            scanf("%d%d",&l,&r);
            printf("%d\n",sum[l][r]);
        }
    }
    return 0;
}