HDU3068(Manacher算法)

Posted 2020-08-13 Penn000

最长回文

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18219    Accepted Submission(s): 6689

Problem Description

给出一个只由小写英文字符a,b,c...y,z组成的字符串S,求S中最长回文串的长度.
回文就是正反读都是一样的字符串,如aba, abba等

 

 

Input

输入有多组case,不超过120组,每组输入为一行小写英文字符a,b,c...y,z组成的字符串S
两组case之间由空行隔开(该空行不用处理)
字符串长度len <= 110000

 

 

Output

每一行一个整数x,对应一组case,表示该组case的字符串中所包含的最长回文长度.

 

 

Sample Input

aaaa

abab

 

 

Sample Output

4

3

 

 

Source

2009 Multi-University Training Contest 16 - Host by NIT

 

 1 //2016.10.06
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 
 6 using namespace std;
 7 
 8 const int N = 110005;
 9 char str[N], s[N<<1];
10 int a[N<<1];
11 
12 int manacher(char *s, int *a, int len)
13 {
14     a[0] = 0;
15     int ans = 0, j;
16     for(int i = 0; i < len; )
17     {
18         while(i-a[i]>0 && s[i+a[i]+1]==s[i-a[i]-1])
19               a[i]++;
20         if(ans < a[i])ans = a[i];
21         j = i+1;
22         while(j<=i+a[i] && i-a[i]!=i+i-j-a[i+i-j]){
23             a[j] = min(a[i+i-j], i+a[i]-j);
24             j++;
25         }
26         a[j] = max(i+a[i]-j, 0);
27         i = j;
28     }
29     return ans;
30 }
31 
32 int main()
33 {
34     int len;
35     while(scanf("%s", str)!=EOF)
36     {
37         len = 2*strlen(str)+1;
38         for(int i = 0; str[i] != ‘\0‘; i++)
39         {
40             s[i+i] = ‘\0‘;
41             s[i+i+1] = str[i];
42         }
43         s[len-1] = ‘\0‘;
44         printf("%d\n", manacher(s, a, len));
45     }
46 
47     return 0;
48 }