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  • Ural 1960 Palindromes and Super Abilities

    Palindromes and Super Abilities

    Time Limit: 1000ms
    Memory Limit: 65536KB
    This problem will be judged on Ural. Original ID: 1960
    64-bit integer IO format: %lld      Java class name: (Any)
    After solving seven problems on Timus Online Judge with a word “palindrome” in the problem name, Misha has got an unusual ability. Now, when he reads a word, he can mentally count the number of unique nonempty substrings of this word that are palindromes.
    Dima wants to test Misha’s new ability. He adds letters s1, ..., sn to a word, letter by letter, and after every letter asks Misha, how many different nonempty palindromes current word contains as substrings. Which n numbers will Misha say, if he will never be wrong?
     

    Input

    The only line of input contains the string s1...sn, where si are small English letters (1 ≤ n ≤ 105).
     

    Output

    Output n numbers separated by whitespaces, i-th of these numbers must be the number of different nonempty substrings of prefix s1...si that are palindromes.
     

    Sample Input

    aba

    Sample Output

    1 2 3

    Source

     
    解题:PalindromicTree
     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 const int maxn = 100010;
     4 struct PalindromicTree{
     5     int fail[maxn],len[maxn],son[maxn][26];
     6     int tot,last,n;
     7     char s[maxn];
     8     int newnode(int slen = 0){
     9         memset(son[tot],0,sizeof son[tot]);
    10         len[tot] = slen;
    11         return tot++;
    12     }
    13     void init(){
    14         n = tot = last = 0;
    15         newnode(0);
    16         newnode(-1);
    17         fail[1] = fail[0] = 1;
    18         s[n] = -1;
    19     }
    20     int getFail(int x){
    21         while(s[n - len[x] - 1] != s[n]) x = fail[x];
    22         return x;
    23     }
    24     void extend(int c){
    25         s[++n] = c;
    26         int cur = getFail(last);
    27         if(!son[cur][c]){
    28             int x = newnode(len[cur] + 2);
    29             fail[x] = son[getFail(fail[cur])][c];
    30             son[cur][c] = x;
    31         }
    32         last = son[cur][c];
    33     }
    34 }pt;
    35 char str[maxn];
    36 int main(){
    37     while(~scanf("%s",str)){
    38         pt.init();
    39         bool flag = false;
    40         for(int i = 0; str[i]; ++i){
    41             pt.extend(str[i] - 'a');
    42             if(flag) putchar(' ');
    43             printf("%d",pt.tot - 2);
    44             flag = true;
    45         }
    46         putchar('
    ');
    47     }
    48     return 0;
    49 }
    View Code
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  • 原文地址:https://www.cnblogs.com/crackpotisback/p/4913902.html
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