HDU 3948 The Number of Palindromes

The Number of Palindromes

Time Limit: 3000ms

Memory Limit: 262144KB

This problem will be judged on HDU. Original ID: 3948
64-bit integer IO format: %I64d      Java class name: Main

Now, you are given a string S. We want to know how many distinct substring of S which is palindrome.

 

Input

The first line of the input contains a single integer T(T<=20), which indicates number of test cases.
Each test case consists of a string S, whose length is less than 100000 and only contains lowercase letters.

 

Output

For every test case, you should output "Case #k:" first in a single line, where k indicates the case number and starts at 1. Then output the number of distinct substring of S which is palindrome.

 

Sample Input

3
aaaa
abab
abcd

Sample Output

Case #1: 4
Case #2: 4
Case #3: 4

Source

2011 Multi-University Training Contest 11 - Host by UESTC

 

解题：Palindromic Tree

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
struct PalindromicTree{
    int son[maxn][26],fail[maxn],len[maxn],s[maxn];
    int tot,last,n;
    int newnode(int slen = 0){
        memset(son[tot],0,sizeof son[tot]);
        len[tot] = slen;
        return tot++;
    }
    void init(){
        tot = last = n = 0;
        newnode(0);
        newnode(-1);
        fail[0] = fail[1] = 1;
        s[n] = -1;
    }
    int getFail(int x){
        while(s[n - len[x] -1] != s[n]) x = fail[x];
        return x;
    }
    void extend(int c){
        s[++n] = c;
        int cur = getFail(last);
        if(!son[cur][c]){
            int x = newnode(len[cur] + 2);
            fail[x] = son[getFail(fail[cur])][c];
            son[cur][c] = x;
        }
        last = son[cur][c];
    }
}pt;
char str[maxn];
int main(){
    int kase,cs = 1;
    scanf("%d",&kase);
    while(kase--){
        scanf("%s",str);
        pt.init();
        for(int i = 0; str[i]; ++i)
            pt.extend(str[i] - 'a');
        printf("Case #%d: %d\n",cs++,pt.tot - 2);
    }
    return 0;
}