HDOJ 3948 The Number of Palindromes 回文串自动机

看上去像是回文串自动机的模板题，就来了一发

The Number of Palindromes

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1992    Accepted Submission(s): 694

Problem Description

Now, you are given a string S. We want to know how many distinct substring of S which is palindrome.

 

Input

The first line of the input contains a single integer T(T<=20), which indicates number of test cases.
Each test case consists of a string S, whose length is less than 100000 and only contains lowercase letters.

 

Output

For every test case, you should output "Case #k:" first in a single line, where k indicates the case number and starts at 1. Then output the number of distinct substring of S which is palindrome.

 

Sample Input

   
   
   
   

    
    
    
    3
aaaa
abab
abcd
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 4
Case #2: 4
Case #3: 4
   
   
   
   

 

Source

2011 Multi-University Training Contest 11 - Host by UESTC

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年04月17日 星期五 09时38分22秒
File Name     :pt.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int C=30;
const int maxn=120000;

struct Node 
{
	Node *ch[C],*suffix;
	int len;
}bar[maxn],*foo,*last,*odd,*even;

char s[maxn];
int n,cnt; 
// cnt=foo-bar=count of palindromes , n the number of char added

void init()
{
	odd=bar; even=last=odd+1; foo=even+1;
	memset(odd->ch,0,sizeof(odd->ch));
	memset(even->ch,0,sizeof(even->ch));
	odd->suffix=even->suffix=odd;
	odd->len=-1; even->len=0;
	n=0; cnt=0;
}

Node* New_Node(int x)
{
	memset(foo->ch,0,sizeof(foo->ch));
	foo->len=x;
	return foo++;
}

int index(char x) { return x-'a'; }

Node* get(Node*p)
{
	while(n-p->len-2<0||s[n-p->len-2]!=s[n-1])
		p=p->suffix;
	return p;
}

bool add(char c)
{
	int x=index(c); s[n++]=c;
	Node* p =get(last);
	if(!p->ch[x])
	{
		last = New_Node(p->len+2);
		if(last->len==1) last->suffix=even;
		else last->suffix=get(p->suffix)->ch[x];
		/// guarantee proper suffix
		p->ch[x]=last; cnt++;
		return true;
	}
	else
	{
		last=p->ch[x];
		return false;
	}
}

char str[maxn];

int main()
{
	int T_T,cas=1;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%s",str);
		int len=strlen(str);
		init();
		for(int i=0;i<len;i++)
		{
			add(str[i]);
		}
		printf("Case #%d: %d\n",cas++,cnt);
	}
}