马拉车用于解决最长回文子串问题，重点是子串，而不是子序列，时间复杂度为O(n)。

解释一下变量的意义：

Len[i]数组去存第i个位置到mx位置的长度

id记录上一次操作的位置（这个操作可以看模板）

mx标记上一次的最长子串的最右端

 

模板：

 1 void init()  //这个是用来处理字符串的
 2 {
 3     memset(str,0,sizeof(str));
 4     int k=0;
 5     str[k++]='$';
 6     for(int i=0;i<len;++i)
 7         str[k++]='#',str[k++]=s[i];
 8     str[k++]='#';
 9     len=k;
10 }
11 int manacher() //求最长回文子串
12 {
13     Len[0]=0;
14     int sum=0;
15     int id,mx=0;
16     for(int i=1;i<len;++i)
17     {
18         if(i<mx)  Len[i]=min(mx-i,Len[2*id-i]);
19         else Len[i]=1;
20         while(str[i-Len[i]]==str[i+Len[i]]) Len[i]++;
21         if(Len[i]+i>mx)
22         {
23             mx=Len[i]+i;
24             id=i;
25             sum=max(sum,Len[i]);
26         }
27     }
28     return (sum-1);
29 }

 

 

当我们要求的以第i个字符为回文字符串的中心的时候，如果i>=mx这个时候没法优化，就是判断(i-1)==(i+1)、(i-2)==(i+2)....一直这样找

看代码就是进行19行、再进行20行

 

如果i<mx的时候，这个时候

这个时候看一道模板题：

 

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

这个时候要注意

不知道是这里memset(Len,0,sizeof(Len)); 导致的超时

还是

 1 void init()
 2 {
 3     memset(str,0,sizeof(str));
 4     int k=0;
 5     str[k++]='$';
 6     for(int i=0;i<strlen(s);++i) 这个strlen导致的
 7         str[k++]='#',str[k++]=s[i];
 8     str[k++]='#';
 9     len=k;
10 }

正确代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<set>
 6 using namespace std;
 7 const int maxn=3000005;
 8 const int INF=0x3f3f3f3f;
 9 const int mod=998244353;
10 char str[maxn],s[maxn];
11 int len,Len[maxn];
12 void init()
13 {
14     memset(str,0,sizeof(str));
15     int k=0;
16     str[k++]='$';
17     for(int i=0;i<len;++i)
18         str[k++]='#',str[k++]=s[i];
19     str[k++]='#';
20     len=k;
21 }
22 int manacher()
23 {
24     Len[0]=0;
25     int sum=0;
26     int id,mx=0;
27     for(int i=1;i<len;++i)
28     {
29         if(i<mx)  Len[i]=min(mx-i,Len[2*id-i]);
30         else Len[i]=1;
31         while(str[i-Len[i]]==str[i+Len[i]]) Len[i]++;
32         if(Len[i]+i>mx)
33         {
34             mx=Len[i]+i;
35             id=i;
36             sum=max(sum,Len[i]);
37         }
38     }
39     return (sum-1);
40 }
41 int main()
42 {
43     int t=0;
44     while(~scanf("%s",s))
45     {
46         //memset(Len,0,sizeof(Len));
47         if(strcmp("END",s)==0) break;
48         len=strlen(s);
49         init();
50         printf("Case %d: %d\n",++t,manacher());
51     }
52     return 0;
53 }

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