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本文作者:ljh2000
作者博客:http://www.cnblogs.com/ljh2000-jump/
转载请注明出处,侵权必究,保留最终解释权!
Description
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
Output
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;
}