马 拉 车 模板题

1.题目引入：

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. 

2.样例输出： 

Sample Input

abcbabcbabcba
abacacbaaaab
END

Sample Output

Case 1: 13
Case 2: 6

这道题是典型的马拉车算法题，这一节的具体解释我们在上一节中已经讲到这里就不过多赘述，有遗忘的可以看一下我们上一节的：https://blog.youkuaiyun.com/weixin_52914088/category_11312023.html  ，搞懂上一节这一节的就就很好理解，具体代码如下：

3.代码如下： 

// 马拉车 
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e6+10;
char s[maxn],s1[maxn*2];
int l[maxn*2],t,len;
void find()
{
	int k=0;
	s1[k++]='&';
	for(int i=0;i<t;i++)
	{
		s1[k++]='#';
		s1[k++]=s[i]; 
	}	
	s1[k++]='#';
	len=k;
} 
int main()
{
	int num=1;
	while(cin>>s)
	{
		if(s[0]=='E') break;
		memset(l,0,sizeof(l));
		t=strlen(s);
		find();
		int id=0,max1=0,ans=0,i;
		for(int i=1;i<len;i++)
		{
			if(ans>i) l[i]=min(ans-i,l[2*id-i]);   // ans 表示右侧边界  l[2*id-i]表示的是 j 的位置回文半径 （i在右侧边界内） 
			else
			{
				l[i]=1;  // 如果i在右侧边界外 起始为 1 向两边扩张 
			}
			while(s1[i+l[i]]==s1[i-l[i]])
			{
				l[i]++;   // 统计的是回文半径 
			}
			if(l[i]+i>ans)
			{
				id=i;      // 记录中心节点的位置 
				ans=l[i]+i;   // 更新右边界 
			}
			max1=max(max1,l[i]-1);  // 原回文长度 等于现回文半径-1 
		}
		cout<<"Case "<<num++<<": "<<max1<<"\n";  // 输出最长回文长度 
	}
	return 0;
}