HDU 5658 CA Loves Palindromic （回文树。）

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

Hint

In first query, the palindromic substrings in the substring $S[1,2]$ are "a","b". In second query, the palindromic substrings in the substring $S[1,2]$ are "a","b","bb". Note that the substring "b" appears twice, but only be counted once. You may need an input-output optimization.

 

Source

BestCoder Round #78 (div.2)

 

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给定一个串，询问l到r右多少本质不同的回文串。

回文树

#include
#include
#include
#include
#include
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))

using namespace std;

typedef long long ll;
const int maxn=1100;
const int INF=1e9+10;

int n,q;
char s[maxn];
int l,r;
ll ans[maxn][maxn];

struct PalinTree
{
    int ch[maxn][26],f[maxn];
    int n,tot,last;
    int len[maxn],cnt[maxn];
    int s[maxn];
    int newnode(int l)
    {
        MS0(ch[tot]);
        cnt[tot]=0;
        len[tot]=l;
        return tot++;
    }
    void init()
    {
        tot=0;
        newnode(0);
        newnode(-1);
        last=0;
        n=0;
        s[n]=-1;
        f[0]=1;
    }
    int get_fail(int x)
    {
        while(s[n-len[x]-1]!=s[n]) x=f[x];
        return x;
    }
    void add(int c)
    {
        c-='a';
        s[++n]=c;
        last=get_fail(last);
        if(!ch[last][c])
        {
            int cur=newnode(len[last]+2);
            f[cur]=ch[get_fail(f[last])][c];
            ch[last][c]=cur;
        }
        last=ch[last][c];
        cnt[last]++;
    }
};
PalinTree pt;

void Init()
{
    int len=strlen(s+1);
    REP(l,1,len)
    {
        pt.init();
        REP(r,l,len)
        {
            pt.add(s[r]);
            ans[l][r]=pt.tot-2;
        }
    }
}

int main()
{
    // freopen("in.txt","r",stdin);
    while(~scanf("%d",&n))
    {
        scanf("%s",s+1);
        Init();
        scanf("%d",&q);
        while(q--)
        {
            scanf("%d%d",&l,&r);
            printf("%I64d\n",ans[l][r]);
        }
    }
    return 0;
}