Palindrome

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;
}

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