HDU5658：CA Loves Palindromic （回文树,求区间本质不同的回文串数）

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.

InputFirst 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
OutputFor each testcase, output the answer in Q lines.Sample Input

1
abba
2
1 2
1 3

Sample Output

2
3

题意：给定字符串S，|S|<1000；Q次询问区间不用本质的回文串数量。

思路：以每个左端点建立回文树即可。 注意判定边界，即给第0位赋值-1啥的，我之前的板子没有，害我wa了几发。

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=;
char c[maxn]; int bb[maxn];
int N,Q,fcy[maxn][maxn];
struct PT
{
    struct node{
        int fail,len,son[];
    }t[maxn];
    int tot,last;
    void init()
    {
        memset(t,,sizeof(t));
        t[].fail=t[].fail=;
        t[].len=-;
        last=; tot=; bb[]=-; bb[]=-;
    }
    void add(int s,int n)
    {
        int p=last; bb[n]=s;
        while(bb[n-t[p].len-]!=bb[n]) p=t[p].fail;
        if(!t[p].son[s])
        {
            int v=++tot,k=t[p].fail;
            t[v].len=t[p].len+;
            while(bb[n-t[k].len-]!=bb[n]) k=t[k].fail;
            t[v].fail=t[k].son[s];
            t[p].son[s]=v;
        }
        last=t[p].son[s];
    }
}T;
void solve()
{
    rep(i,,N){
        T.init();
        rep(j,i,N) {
            T.add(c[j]-'a',j-i+);
            fcy[i][j]=T.tot-;
        }
    }
    scanf("%d",&Q);
    rep(i,,Q){
       int L,R; scanf("%d%d",&L,&R);
       printf("%d\n",fcy[L][R]);
    }
}
int main()
{
    int T;scanf("%d",&T);
    while(T--){
        memset(c,,sizeof(c));
        scanf("%s",c+);
        N=strlen(c+);
        solve();
    }
    return ;
}