HDU 5658 CA Loves Palindromic（回文树）

CA Loves Palindromic

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 301    Accepted Submission(s): 131

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

1abba21 21 3

 

Sample Output

23
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;}