hdu 3068 最长回文 (manacher模板)

来源：互联网发布：淘宝东西怎么上架编辑：程序博客网时间：2024/06/05 15:03

最长回文

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9421 Accepted Submission(s): 3240

Problem Description

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

aaaaabab

Sample Output

Source

2009 Multi-University Training Contest 16 - Host by NIT

Recommend

lcy

题目大意:求最长回文子串
题目分析:manacher模板,不解释

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#define MAX 110007using namespace std;char s[MAX];char temp[MAX<<1];int Len[MAX<<1];int init ( char *st ){    int i , len = strlen(st);    temp[0]='@';    for ( i = 1; i <= 2*len; i += 2 )    {        temp[i] = '#';        temp[i+1] = st[i/2];    }    temp[2*len+1] = '#';    temp[2*len+2] = '$';    temp[2*len+3] = 0;    return 2*len+1;}int manacher ( char *st , int len ){    int mx = 0 , ans = 0 , po = 0;    for ( int i = 1 ; i <= len; i++ )    {        if ( mx > i )            Len[i] = min ( mx - i , Len[2*po-i] );        else             Len[i] = 1;        while ( st[i-Len[i]] == st[i+Len[i]] )            Len[i]++;        if ( Len[i]+i > mx )            mx = Len[i]+i , po = i;        ans = max ( ans , Len[i] );    }    return ans-1;}int main ( ){    while ( ~scanf ( "%s" , s ) )    {        printf ( "%d\n" , manacher ( temp , init ( s ) ) );    }}

0 0