hdu 3068 最长回文(manachar模板)

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

aaaa

 

abab

 

Sample Output

4

3

 

Source

2009 Multi-University Training Contest 16 - Host by NIT

 

Recommend

lcy   |   We have carefully selected several similar problems for you:  1358 3336 3065 3746 1711 

 

manachar讲解:   http://blog.csdn.net/bruce_zeng/article/details/8629572

代码:

#include <vector>
#include <set>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <cstring>
#include <queue>
#include <stack>
using namespace std;
#define N 110010*2

char s[N],str[N];
int p[N];

int init()
{
    int i,j=0;
    str[j++]='$';
    for(i=0;s[i];i++){
        str[j++]='#';
        str[j++]=s[i];
    }
    str[j++]='#';
    str[j]='';
    return j;
}

void manachar(int n)
{
    int mx=0,id,i;
    p[0]=0;
    for(i=1;i<n;i++){
        if(mx>i)
            p[i]=min(mx-i,p[2*id-i]);
        else
            p[i]=1;
        while(str[i-p[i]]==str[i+p[i]])
            p[i]++;
        if(p[i]+i>mx){
            mx=p[i]+i;
            id=i;
        }
    }
}

int main()
{
    int n;
    while(~scanf("%s",s)){
        n=init();
        manachar(n);
        int ans=0;
        for(int i=1;i<n; i++){
            if(p[i]>ans)
                ans=p[i];
        }
        printf("%d
",ans-1);
    }
}