URAL 2040 Palindromes and Super Abilities 2(回文树)

Palindromes and Super Abilities 2

Time Limit: 1MS

Memory Limit: 102400KB

64bit IO Format: %I64d & %I64u

Status

Description

Dima adds letters s₁, …, 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 nnumbers will be said by Misha if it is known that he is never wrong?

Input

The input contains a string s₁ … 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 s₁ … s_i minus the number of palindrome substrings of the
prefixs₁ … 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).

Source

Problem Author: Mikhail Rubinchik (prepared by Kirill Borozdin) 

Problem Source: Ural FU Dandelion contest. Petrozavodsk training camp. Summer 2014 

每插入一个字符都会有两种情况，产生新的回文树节点，不产生新的节点。所以答案只会是0,1

#include <iostream>
#include <string.h>

#include <stdlib.h>
#include <math.h>
#include <stdio.h>

using namespace std;
typedef long long int LL;
const int MAX=5*1e6;
const int maxn=4*1e6+5;
char str[MAX+5];
struct Tree
{
    int next[maxn][2];
    int fail[MAX+5];
    int len[MAX+5];
    int s[MAX+5];
    int last,n,p;
    int new_node(int x)
    {
        memset(next[p],0,sizeof(next[p]));
        len[p]=x;
        return p++;
    }
    void init()
    {
        p=0;
        new_node(0);
        new_node(-1);
        last=0;n=0;
        s[0]=-1;
        fail[0]=1;
	}
    int get_fail(int x)
    {
        while(s[n-len[x]-1]!=s[n])
            x=fail[x];
        return x;
    }
    int add(int x)
    {
        x-='a';
        s[++n]=x;
        int cur=get_fail(last);
        if(!(last=next[cur][x]))
        {
            int now=new_node(len[cur]+2);
            fail[now]=next[get_fail(fail[cur])][x];
            next[cur][x]=now;
            last=now;
            return 1;
        }
        return 0;
    }

}tree;
char ans[MAX+5];
int main()
{
    while(scanf("%s",str)!=EOF)
    {

        tree.init();
		int i;
        for( i=0;str[i];i++)
        {
			if(!tree.add(str[i]))
			    ans[i]='0';
			else
				ans[i]='1';
		}
		ans[i]='\0';
		puts(ans);
	}
    return 0;
}