Palindromes and Super Abilities 2
Time Limit: 1MS | Memory Limit: 102400KB | 64bit IO Format: %I64d & %I64u |
Description
Dima adds letters s1, …, sn 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 … sn 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 … si minus the number of palindrome substrings of the
prefixs1 … si?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; }