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): 374 Accepted Submission(s): 161

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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回文树。给出的串总长度只有1000，问任意区间中本质不同的回文串数量。因为长度才1000就暴力枚举区间即可。

#include "cstring"#include "cstdio"#include "string.h"#include "iostream"using namespace std;const int MAXN = 1005 ;const int N = 26 ;char str[1005];struct Palindromic_Tree {    int next[MAXN][N] ;//next指针，next指针和字典树类似，指向的串为当前串两端加上同一个字符构成    int fail[MAXN] ;//fail指针，失配后跳转到fail指针指向的节点    int cnt[MAXN] ;    int num[MAXN] ;    int len[MAXN] ;//len[i]表示节点i表示的回文串的长度    int S[MAXN] ;//存放添加的字符    int last ;//指向上一个字符所在的节点，方便下一次add    int n ;//字符数组指针    int p ;//节点指针    int newnode ( int l ) {//新建节点        for ( int i = 0 ; i < N ; ++ i ) next[p][i] = 0 ;        cnt[p] = 0 ;        num[p] = 0 ;        len[p] = l ;        return p ++ ;    }    void init () {//初始化        p = 0 ;        newnode (  0 ) ;        newnode ( -1 ) ;        last = 0 ;        n = 0 ;        S[n] = -1 ;//开头放一个字符集中没有的字符，减少特判        fail[0] = 1 ;    }    int get_fail ( int x ) {//和KMP一样，失配后找一个尽量最长的        while ( S[n - len[x] - 1] != S[n] ) x = fail[x] ;        return x ;    }    void add ( int c ) {        c -= 'a' ;        S[++ n] = c ;        int cur = get_fail ( last ) ;//通过上一个回文串找这个回文串的匹配位置        if ( !next[cur][c] ) {//如果这个回文串没有出现过，说明出现了一个新的本质不同的回文串            int now = newnode ( len[cur] + 2 ) ;//新建节点            fail[now] = next[get_fail ( fail[cur] )][c] ;//和AC自动机一样建立fail指针，以便失配后跳转            next[cur][c] = now ;            num[now] = num[fail[now]] + 1 ;        }        last = next[cur][c] ;        cnt[last] ++ ;    }    void count () {        for ( int i = p - 1 ; i >= 0 ; -- i ) cnt[fail[i]] += cnt[i] ;        //父亲累加儿子的cnt，因为如果fail[v]=u，则u一定是v的子回文串！    }}tree;int main(){    int cas;    scanf("%d",&cas);    while(cas--)    {        long long ans[1005][1005];        //Palindromic_Tree tree;        scanf("%s",str+1);        tree.init();        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]);                ans[i][j]=tree.p-2;            }        }        int cnt;        scanf("%d",&cnt);        while(cnt--)        {            int l,r;            scanf("%d%d",&l,&r);            printf("%lld\n",ans[l][r]);        }    }}