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)

Recommend

wange2014 | We have carefully selected several similar problems for you: 5910 5909 5908 5907 5906

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]
;//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
= -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
) 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]);
}

}
}