题目描述:
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.
输入:
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).
输出:
For each test case in the input print the test case number and the length of the largest palindrome.
样例输入:
abcbabcbabcba
abacacbaaaab
END
样例输出:
Case 1: 13
Case 2: 6
code:
题意:求最长的回文串
思路:Manacher算法
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define M 1000010
char a[M];
char b[2*M];
int feng[2*M];
void init()
{
memset(feng,0,sizeof(feng));
int j=0;
b[j++]='$';
b[j++]='#';
int n=strlen(a);
for(int i=0;i<n;i++)
{
b[j++]=a[i];
b[j++]='#';
}
b[j]='\0';
}
void manacher(int cas)
{
init();
int n=strlen(b);
int id=-1,mx=-1,maxx=0;
for(int i=1;i<n;i++)
{
if(i<mx)
{
feng[i]=min(mx-i,feng[id*2-i]);
}
else feng[i]=1;
while(b[i+feng[i]]==b[i-feng[i]])feng[i]++;
if(i+feng[i]>mx)
{
mx=i+feng[i];
id=i;
}
maxx=max(maxx,feng[i]);
}
printf("Case %d: %d\n",cas,maxx-1);
}
int main()
{
int kk=1;
while(~scanf("%s",a))
{
if(!strcmp(a,"END"))
break;
manacher(kk++);
}
return 0;
}