HDU 3068 最长回文

最长回文

Time Limit: 2000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 3068
64-bit integer IO format: %I64d      Java class name: Main

给出一个只由小写英文字符a,b,c...y,z组成的字符串S,求S中最长回文串的长度.
回文就是正反读都是一样的字符串,如aba, abba等

 

Input

输入有多组case,不超过120组,每组输入为一行小写英文字符a,b,c...y,z组成的字符串S
两组case之间由空行隔开(该空行不用处理)
字符串长度len <= 110000

 

Output

每一行一个整数x,对应一组case,表示该组case的字符串中所包含的最长回文长度.

 

Sample Input

aaaa

abab

Sample Output

4
3

Source

2009 Multi-University Training Contest 16 - Host by NIT

 

解题：Palindromic Tree

#include <bits/stdc++.h>
using namespace std;
const int maxn = 200010;
struct PalindromicTree{
    int son[maxn][26],fail[maxn],len[maxn],s[maxn];
    int tot,last,n,maxlen;
    int newnode(int slen = 0){
        len[tot] = slen;
        memset(son[tot],0,sizeof son[tot]);
        return tot++;
    }
    void init(){
        maxlen = 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;
            maxlen = max(maxlen,len[x]);
        }
        last = son[cur][c];
    }
}pt;
char str[maxn];
int main(){
    while(~scanf("%s",str)){
        pt.init();
        for(int i = 0; str[i]; ++i)
            pt.extend(str[i] - 'a');
        printf("%d
",pt.maxlen);
    }
    return 0;
}