HDU 5394 Trie in Tina Town

Problem Description

Tina Town is a friendly place. People there care about each other.

A trie which was planted by the first mayor of Tina Town grows in the center of the town.

We define a palindrome substring in a trie a string that is a suffix of a string which the path from the root to any node represents and the string is a palindrome. Two palindromes are different if and only if their positions are different. Now, Tina wants to know the sum of the length of all palindrome substrings. Tina didn’t know the answer so she asked you to find out the answer for her.

 

Input

The first line contains a integer – number of cases
For each case, the first line is an integer  N  representing the number of nodes in trie except the root.
The following  N  lines contains a letter between  a  and  d  – the letter that node  n[i]  stores and a number  f[i]  – the index of  n[i] ’s father. It’s guaranteed that  fa[i]≤i . If  fa[i]=0  n[i] is the root.
T≤10,N≤2∗106

 

Output

The first line contains a integer – number of cases
For each case, the first line is an integer  N  representing the number of nodes in trie except the root.
The following  N  lines contains a letter between  a  and  d  – the letter that node  n[i]  stores and a number  f[i]  – the index of  n[i] ’s father. It’s guaranteed that  fa[i]≤i . If  fa[i]=0  n[i] is the root.
T≤10,N≤2∗106

 

Sample Input

   
   
   
   

    
    
    
    2
5
a 0
a 1
a 2
b 1
b 4
5
a 0
a 1
a 2
b 1
a 4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    14 15 
    
    
    
    
 
     
 
      Hint 
     
 The first test case is a trie like this: aaa *1 -> 3 aa *2 -> 4 a *3-> 3 b *2-> 2 bb *1- > 2 3+4+3+2+2 = 14 The second test case: aaa *1 -> 3 aba *1 -> 3 a *4 -> 4 b *1 -> 1 aa *2 -> 4 3+3+4+1+4 = 15 If the stack size is too small, you can submit it in C++ and add ‘#pragma comment(linker, "/STACK:102400000,102400000”)’ at the head of your program. Large input, recommend to use fast I/O. 
    
    
    
    
    
 
    
    
    
    
回文树的应用，这里的操作有加入一个字母和删除一个字母，以及计算一个前缀对应的全部后缀回文串长度和，可以直接在回文树上统计。
    
    
    
    

    
    
    
    
#pragma comment(linker, "/STACK:102400000,102400000")
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int maxn = 2e6 + 10;
int T, n, ft[maxn], nt[maxn], x;
char s[maxn], ch;
LL ans;

struct PalindromicTree
{
	const static int maxn = 2e6 + 10;
	const static int size = 4;
	int next[maxn][size], sz, tot;
	int fail[maxn], len[maxn], last[maxn], pre[maxn];
	LL cnt[maxn];
	char s[maxn];
	void clear()
	{
		len[1] = -1; len[2] = 0;
		fail[1] = fail[2] = 1;
		cnt[1] = cnt[2] = tot = 0;
		last[0] = (sz = 3) - 1;
		memset(next[1], 0, sizeof(next[1]));
		memset(next[2], 0, sizeof(next[2]));
	}
	int Node(int length)
	{
		memset(next[sz], 0, sizeof(next[sz]));
		len[sz] = length;      return sz;
	}
	int getfail(int x)
	{
		while (s[tot] != s[tot - len[x] - 1]) x = fail[x];
		return x;
	}
	LL add(char pos)
	{
		int x = (s[++tot] = pos) - 'a', y = getfail(last[tot - 1]);
		if (next[y][x]) { last[tot] = next[y][x]; pre[tot] = 0; }
		else {
			last[tot] = next[y][x] = Node(len[y] + 2);
			pre[tot] = y;
			fail[sz] = len[sz] == 1 ? 2 : next[getfail(fail[y])][x];
			cnt[sz] = cnt[fail[sz]] + len[sz];        ++sz;
		}
		return cnt[last[tot]];
	}
	void del(char pos)
	{
		int x = pos - 'a';
		if (pre[tot])
		{
			next[pre[tot]][x] = 0;
			--sz;
		}
		--tot;
	}
}solve;

inline int get()
{
	while ((ch = getchar()) < '0' || ch > '9');
	int x = ch - '0';
	while ((ch = getchar()) >= '0' && ch <= '9') x = x * 10 + ch - '0';
	return x;
}

void dfs(int x)
{
	for (int i = ft[x]; i != -1; i = nt[i])
	{
		ans += solve.add(s[i]);
		dfs(i);
		solve.del(s[i]);
	}
}

int main()
{
	scanf("%d", &T);
	while (T--)
	{
		solve.clear();
		scanf("%d", &n);
		for (int i = 0; i <= n; i++) ft[i] = -1;
		for (int i = 1; i <= n; i++)
		{
			//while ((s[i] = getchar()) < 'a' || ch > 'd');
			//x = get(); nt[i] = ft[x];    ft[x] = i;
			scanf("%s%d", s + i, &x);
			nt[i] = ft[x];    ft[x] = i;
		}
		dfs(ans = 0);
		printf("%lld\n", ans);
	}
	return 0;
}