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 = ;
const int N = ;
char str[];
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 = ; i < N ; ++ i ) next[p][i] = ;
cnt[p] = ;
num[p] = ;
len[p] = l ;
return p ++ ;
}
void init () {//初始化
p = ;
newnode ( ) ;
newnode ( - ) ;
last = ;
n = ;
S[n] = - ;//开头放一个字符集中没有的字符,减少特判
fail[] = ;
}
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] - ] != 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] + ) ;//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]] + ;
}
last = next[cur][c] ;
cnt[last] ++ ;
}
void count () {
for ( int i = p - ; i >= ; -- 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[][];
//Palindromic_Tree tree;
scanf("%s",str+);
tree.init();
int len=strlen(str+);
for(int i=;i<=len;i++)
{
tree.init();
for(int j=i;j<=len;j++)
{
tree.add(str[j]);
ans[i][j]=tree.p-;
}
}
int cnt;
scanf("%d",&cnt);
while(cnt--)
{
int l,r;
scanf("%d%d",&l,&r);
printf("%lld\n",ans[l][r]);
}
}
}