ural1960Palindromes and Super Abilities（回文树）

回文树根据以下博客学的，写得很好

传送门：http://blog.csdn.net/u013368721/article/details/42100363

After solving seven problems on Timus Online Judge with a word “palindrome” in the problem name, Misha has got an unusual ability. Now, when he reads a word, he can mentally count the number of unique nonempty substrings of this word that are palindromes.

Dima wants to test Misha’s new ability. He adds letters  s 1, ...,  s n to a word, letter by letter, and after every letter asks Misha, how many different nonempty palindromes current word contains as substrings. Which  n numbers will Misha say, if he will never be wrong?

Input

The only line of input contains the string  s 1... s n, where  s i are small English letters (1 ≤  n ≤ 10 5).

Output

Output  n numbers separated by whitespaces,  i-th of these numbers must be the number of different nonempty substrings of prefix  s 1... s i that are palindromes.

Sample

input	output
aba	1 2 3

题意：大概就是每个前缀有多少种回文串

思路：如果看懂了回文树，就会知道没当开新结点证明出现了新的回文串。

#include 
#include 
#include 
using namespace std;
#define maxn 100080
#define N 26
char str[maxn];
struct Palidromic_Tree
{
	int next[maxn][N];//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;
		fail[0] = 1;
	}

	int get_fail( int x )//和KMP一样，适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;
		//printf("S[%d]=:%d\n",n,S[n]);
		int cur = get_fail( last );//通过上一个回文串找这个回文串的匹配位置
		//printf("cur=%d\n",cur);
		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
			//printf("now=%d\n",now);
			fail[now] = next[get_fail(fail[cur])][c];//和AC自动机一样建立fail指针，以便适配后跳转
			//printf("fail[now]=%d\n",fail[now]);
			next[cur][c] = now;
			//printf("next[%d][%d]=%d\n",cur,c,now);
			num[now] = num[fail[now]] + 1;
			//printf("num[%d]=%d\n",now,num[now]);
		}
		last = next[cur][c];
		//printf("last=%d\n",next[cur][c]);
		cnt[last]++;
		//printf("cnt[last]=%d\n",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
	}
}pt;
int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);
	pt.init();
	scanf("%s",str);
	int len = strlen(str);
	//printf("%s",str);
	for(int i = 0;i < len;i++)
	{
		//for(int j = 0;j < 3;j++)	puts("");
		pt.add(str[i]);
		printf("%d",pt.p-2);
		if(i == len-1)	puts("");
		else printf(" ");
	}
	return 0;
}