URAL 2040 Palindromes and Super Abilities 2

Description

Dima adds letters 

s

1, …, 

s

n one by one to the end of a word. After each letter, he asks Misha to tell him how many new palindrome substrings appeared when he added that letter. Two substrings are considered distinct if they are different as strings. Which 

n numbers will be said by Misha if it is known that he is never wrong?

Input

The input contains a string 
    s
    1 … 
    s
    n consisting of letters ‘a’ and ‘b’ (1 ≤ 
    n ≤ 5 000 000).      

Output

Print 

n numbers without spaces: 

i-th number must be the number of palindrome substrings of the prefix 

s

1 … 

s

i minus the number of palindrome substrings of the prefix 

s

1 … 

s

i−1. The first number in the output should be one.

Sample Input

input	output
abbbba	111111

Notes

We guarantee that jury has C++ solution which fits Time Limit at least two times. We do not guarantee that solution on other languages exists (even Java).

简单回文树应用，然而会卡输出，简直有毒

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
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 = 5e6 + 10;
int T, n;
char s[maxn], ans[maxn];

struct PalindromicTree
{
  const static int maxn = 5e6 + 10;
  const static int size = 2;
  int next[maxn][size], last, sz, tot;
  int fail[maxn], len[maxn];
  char s[maxn];
  void clear()
  {
    len[1] = -1; len[2] = 0;
    fail[2] = fail[1] = 1;
    last = (sz = 3) - 2;
    tot = 0;
    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;
  }
  int add(char pos)
  {
    int x = (s[++tot] = pos) - 'a', y = getfail(last);
    if (next[y][x]) { last = next[y][x]; return 0; }

    last = next[y][x] = Node(len[y] + 2);
    fail[last] = len[last] == 1 ? 2 : next[getfail(fail[y])][x];
    return 1;
  }
}solve;

int main()
{
  while (scanf("%s", s) != EOF)
  {
    solve.clear();
    for (int i = 0; s[i]; i++)
    {
      ans[i] = solve.add(s[i]) + '0';
      ans[i + 1] = 0;
    }
    puts(ans);
  }
  return 0;
}