地址：http://acm.hdu.edu.cn/showproblem.php?pid=4763

题目：

Theme Section

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3491    Accepted Submission(s): 1623

Problem Description

It‘s time for music! A lot of popular musicians are invited to join us in the music festival. Each of them will play one of their representative songs. To make the programs more interesting and challenging, the hosts are going to add some constraints to the rhythm of the songs, i.e., each song is required to have a ‘theme section‘. The theme section shall be played at the beginning, the middle, and the end of each song. More specifically, given a theme section E, the song will be in the format of ‘EAEBE‘, where section A and section B could have arbitrary number of notes. Note that there are 26 types of notes, denoted by lower case letters ‘a‘ - ‘z‘.

To get well prepared for the festival, the hosts want to know the maximum possible length of the theme section of each song. Can you help us?

Input

The integer N in the first line denotes the total number of songs in the festival. Each of the following N lines consists of one string, indicating the notes of the i-th (1 <= i <= N) song. The length of the string will not exceed 10^6.

Output

There will be N lines in the output, where the i-th line denotes the maximum possible length of the theme section of the i-th song.

Sample Input

5
xy
abc
aaa
aaaaba
aaxoaaaaa

Sample Output

0
0
1
1
2

Source

2013 ACM/ICPC Asia Regional Changchun Online

思路：用所给字符串去匹配字符串EAEBE，AB可以为空

　　前缀为E，后缀为E，所以很容易想到kmp的next数组（前缀和后缀的匹配情况），然后查找所有所有前缀匹配后缀的情况然后check一下即可。

　　查找所有前缀和后缀匹配的情况时可以利用next进行转移。

　　ans=next[ans-1]。

 1 #include <bits/stdc++.h>
 2 #define PB push_back
 3 #define MP make_pair
 4 using namespace std;
 5 typedef long long LL;
 6 typedef pair<int,int> PII;
 7 #define PI acos((double)-1)
 8 #define E exp(double(1))
 9 const int K=1e6+9;
10 int nt[K];
11 char a[K];
12
13 //参数为模板串和next数组
14 //字符串均从下标0开始
15 void kmp_next(char *T,int *nt)
16 {
17     nt[0]=0;
18     for(int i=1,j=0,m=strlen(T);i<m;i++)
19     {
20         while(j&&T[i]!=T[j])j=nt[j-1];
21         if(T[i]==T[j])j++;
22         nt[i]=j;
23     }
24 }
25
26 int main(void)
27 {
28     int t;cin>>t;
29     while(t--)
30     {
31         scanf("%s",a);
32         kmp_next(a,nt);
33         int len=strlen(a);
34         int ans=nt[len-1],ff=1;
35         while(ans)
36         {
37             for(int i=ans;i<=len-ans&&ff;i++)
38             if(nt[i]==ans)
39                 ff=0;
40             if(!ff) break;
41             ans=nt[ans-1];
42         }
43         printf("%d\n",ans);
44     }
45
46     return 0;
47 }