HDU4763 Theme Section —— KMP next数组

题目链接：https://vjudge.net/problem/HDU-4763

Theme Section

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4264    Accepted Submission(s): 2056

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

 

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题解：

给定一个字符串，找出能构成"EAEBE"形式的字符串的E的最长长度。其中A和B任意。

1.可知next[len]就是：即是前缀又是后缀的最长子串。所以我们就解决的首尾两个E（如果大于三分之一长度，可以通过next数组回退）。

2.剩下的问题就是：在两个E之间的连续子串中，是否能找到第三个E。

3.怎么找到？枚举里面的每个位置（注意不能与首尾的E有重叠），通过next数组的回退，一直找。详情还是看代码吧。

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <string>
 #include <vector>
 #include <map>
 #include <set>
 #include <queue>
 #include <sstream>
 #include <algorithm>
 using namespace std;
 typedef long long LL;
 const double eps = 1e-;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = 1e6+;

 char x[MAXN];
 int Next[MAXN];

 void get_next(char x[], int m)
 {
     int i, j;
     j = Next[] = -;
     i = ;
     while(i<m)
     {
         while(j!=- && x[i]!=x[j]) j = Next[j];
         Next[++i] = ++j;
     }
 }

 int main()
 {
     int T;
     scanf("%d", &T);
     while(T--)
     {
         scanf("%s", x);
         int len = strlen(x);
         get_next(x, len);
         int r = Next[len];
         while(r> && len/r<) r = Next[r];

         bool hav_ans = false;
         for(; Next[r]>=; r = Next[r])
         {
             for(int i = *r; i<=len-r; i++)
             {
                 int k = i;  //不要写成Next[i];
                 while(Next[k]>r) k = Next[k];
                 if(Next[k]==r)
                 {
                     hav_ans = true;
                     break;
                 }
             }
             if(hav_ans) break;
         }
         printf("%d\n", r);
     }
 }