ural 1960 Palindromes and Super Abilities

题意：找出s[1..x]的本质不同的回文串的个数

分析：回文树模板题

/************************************************
Author        :DarkTong
Created Time  :2016年09月21日 星期三 21时17分07秒
File Name     :a.cpp
 *************************************************/

#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ULL;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double eps = 1e-9;
const int maxn = 100000 + 10;
const int N = 26;

struct Palindromic_Tree
{
	int next[maxn][N];
	int fail[maxn];
	int cnt[maxn];
	int num[maxn];
	int len[maxn];
	int S[maxn];
	int last;
	int n;
	int p;
	int ans[maxn];

	int newNode(int l)
	{
		for(int i=0;i<N;++i) next[p][i]=0;
		cnt[p] = 0;
		num[p] = 0;
		len[p] = l;
		return p ++;
	}

	void init()
	{
		p = 0;
		newNode(0);
		newNode(-1);
		last = 0;
		n = 0;
		S[n] = -1;
		fail[0] = 1;

		ans[0] = 0;
	}

	int get_fail(int x)
	{
		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);
		if(!next[cur][c])
		{
			int now = newNode(len[cur] + 2);
			fail[now] = next[get_fail(fail[cur])][c];
			next[cur][c] = now;
			num[now] = num[fail[now]] + 1;
			ans[n] = ans[n-1] + 1;
		}
		else ans[n] = ans[n-1];
		last = next[cur][c];
		cnt[last] ++;
	}

	void count()
	{
		for(int i=p-1;i>=0;--i) cnt[fail[i]] += cnt[i];
	}

	void solve(char *s)
	{
		int cur = 0;
		int n = strlen(s+1);
		for(int i=1; i<=n ; ++i)
		{
			cur = next[cur][s[i]-‘a‘];
			printf("%d%c", cnt[i], i==n ? ‘\n‘ : ‘ ‘);
		}
	}
}plt;
char s[maxn];

int main()
{
	int T, cas=1;
	scanf("%s", s+1);
	plt.init();
	for(int i=1; s[i] ; ++ i)
		plt.add(s[i]);
	//	plt.count();
	//	plt.solve(s);
	int n = strlen(s+1);
	for(int i=1;i<=n;++i) printf("%d%c", plt.ans[i], i==n ? ‘\n‘ : ‘ ‘);

	return 0;
}