HDU 3068 最长回文

Posted 2020-06-24 月夜下

最长回文

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13853    Accepted Submission(s): 5045

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

 

 

 

解析：manacher算法。

 

 

 

 1 #include <cstdio>
 2 #include <algorithm>
 3 using namespace std;
 4 
 5 char oris[110005];
 6 char news[220005];
 7 int p[220005];
 8 
 9 void pretreatment()
10 {
11     news[0] = ‘$‘;
12     news[1] = ‘#‘;
13     int i,l;
14     for(i = 0,l = 2; oris[i] != ‘\0‘; ++i,l += 2){
15         news[l] = oris[i];
16         news[l+1] = ‘#‘;
17     }
18     news[l] = ‘\0‘;
19 }
20 
21 int manacher()
22 {
23     int idx,mx = 0;
24     int ans = 0;
25     for(int i = 1; news[i] != ‘\0‘; ++i){
26         p[i] = mx>i ? min(p[(idx<<1)-i],mx-i) : 1;
27         while(news[i+p[i]] == news[i-p[i]])
28             ++p[i];
29         if(i+p[i]>mx){
30             mx = i+p[i];
31             idx = i;
32         }
33         if(p[i]>ans)
34             ans = p[i];
35     }
36     return --ans;
37 }
38 
39 int main()
40 {
41     while(~scanf("%s",oris)){
42         pretreatment();
43         printf("%d\n",manacher());
44     }
45     return 0;
46 }