POJ3974 Palindrome

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本文作者：ljh2000
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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.

Input

Your program will be tested on at most 30 test cases, each test case is given as a string of at most 1000000 lowercase characters on a line by itself. The input is terminated by a line that starts with the string "END" (quotes for clarity). 

Output

For each test case in the input print the test case number and the length of the largest palindrome. 

Sample Input

abcbabcbabcba
abacacbaaaab
END

Sample Output

Case 1: 13
Case 2: 6

正解：manacher

解题报告：

　　manacher模板题。

　　网上有一篇博客写得很清楚：https://segmentfault.com/a/1190000003914228。

　　大致做法就是：维护一个目前所接触的最右端（不妨设为$maxRight$），和使得最长回文子串取到最右端的对称中心位置（不妨设为$id$），我们需要对于每个i求$p[i]$，$p[i]$表示以$i$为中心的最长回文串半径（包括$i$自身）。

　　每次处理i时，分类讨论：

　　如果$i>=maxRight$，显然$i$只能重新开始拓展，不能利用前面的结果，暴力比较即可；

　　如果$i<maxRight$，又可以分两种情况讨论，那篇博客里面说的很清楚了，分为$p[j]$很长和$p[j]$很短两种情况——总之就是可以利用之前的对于现在的$p[i]$确定一个初值，即$p[i]=min(p[2*id-i],maxRight-i)$。

　　容易发现最后$max(p[i]-1)$就是我们想求的答案。

//It is made by ljh2000
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long LL;
const int MAXN = 2000011;
const int MAXM = 1000011;
int len,n,p[MAXN],Case,ans,id,maxRight;
char ch[MAXM],s[MAXN];

inline int getint(){
    int w=0,q=0; char c=getchar(); while((c<'0'||c>'9') && c!='-') c=getchar();
    if(c=='-') q=1,c=getchar(); while (c>='0'&&c<='9') w=w*10+c-'0',c=getchar(); return q?-w:w;
}

inline void manacher(){
	s[0]='%'; s[1]='#'; n=1;
	for(int i=0;i<len;i++) {
		s[++n]=ch[i];
		s[++n]='#';//插入字符，化偶为奇
	}
	maxRight=0; id=0;
	for(int i=1;i<=n;i++) {
		//p[i]表示以i为中心的最长回文串半径（包括i）
		if(i<maxRight) p[i]=min(p[2*id-i],maxRight-i); else p[i]=1;
		for(;i+p[i]<=n && s[i-p[i]]==s[i+p[i]];p[i]++) ;
		if(i+p[i]>maxRight) { id=i; maxRight=i+p[i]; }
	}
	for(int i=1;i<=n;i++) ans=max(ans,p[i]);
	ans--;
}

inline void work(){
	while(scanf("%s",ch)!=EOF) {
		len=strlen(ch);	if(len==3 && ch[0]=='E' && ch[1]=='N' && ch[2]=='D') break;
		memset(p,0,sizeof(p)); ans=1;
		manacher();	Case++;
		printf("Case %d: %d\n",Case,ans);
	}
}

int main()
{
    work();
    return 0;
}