POJ 3974 Palindrome

最新推荐文章于 2022-10-21 15:25:37 发布

原创最新推荐文章于 2022-10-21 15:25:37 发布 · 274 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#poj

Manacher 专栏收录该内容

11 篇文章

订阅专栏

本文介绍了一种高效的算法——Manacher算法，用于解决寻找字符串中最长回文子串的问题。通过实例说明了该算法的基本思想及实现过程，并提供了一个完整的C++代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description
Andy the smart computer science student was attending an algorithms class when the professor asked the students a simple question, “Can you propose an efficient algorithm to find the length of the largest palindrome in a string?”
A string is said to be a palindrome if it reads the same both forwards and backwards, for example “madam” is a palindrome while “acm” is not.
The students recognized that this is a classical problem but couldn’t come up with a solution better than iterating over all substrings and checking whether they are palindrome or not, obviously this algorithm is not efficient at all, after a while Andy raised his hand and said “Okay, I’ve a better algorithm” and before he starts to explain his idea he stopped for a moment and then said “Well, I’ve an even better algorithm!”.
If you think you know Andy’s final solution then prove it! Given a string of at most 1000000 characters find and print the length of the largest palindrome inside this string.

【题目分析】
manacher算法跑一边，记录最大值，输出就可以了。

【代码】

#include <cstdio>
#include <cstring>
using namespace std;
char s[1010000],ss[2010000];
int r[2010000],kase;
inline int min(int a,int b)
{return a>b?b:a;}
int main()
{
    while (scanf("%s",s+1)!=EOF&&s[1]!='E')
    {
        memset(r,0,sizeof r);
        int l=strlen(s+1);
        ss[1]='#';ss[0]='@';
        for (int i=1;i<=l;++i){
            ss[i*2]=s[i];
            ss[i*2+1]='#';
        }
        s[l*2+2]='$';
        l=l*2+1;
        int id=0,mx=0,ans=0,maxi=0;
        for (int i=1;i<=l;++i){
            if (mx>i) r[i]=min(r[2*id-i],mx-i); else r[i]=0;
            while (ss[i-r[i]]==ss[i+r[i]]) r[i]++;
            if (i+r[i]>mx) mx=i+r[i],id=i;
            if (r[i]>maxi) maxi=r[i];
        }
        printf("Case %d: %d\n",++kase,(maxi*2-1)/2);
    }
}