KMP循环节

链接：https://www.nowcoder.com/acm/contest/141/E

题目描述

Eddy likes to play with string which is a sequence of characters. One day, Eddy has played with a string S for a long time and wonders how could make it more enjoyable. Eddy comes up with following procedure:

1. For each i in [0,|S|-1], let S _i be the substring of S starting from i-th character to the end followed by the substring of first i characters of S. Index of string starts from 0.
2. Group up all the S _i. S _i and S _j will be the same group if and only if S _i=S _j.
3. For each group, let L _j be the list of index i in non-decreasing order of S _i in this group.
4. Sort all the L _j by lexicographical order.

Eddy can't find any efficient way to compute the final result. As one of his best friend, you come to help him compute the answer!

输入描述:

Input contains only one line consisting of a string S.

1≤ |S|≤ 10

⁶

S only contains lowercase English letters(i.e. 

).

输出描述:

First, output one line containing an integer K indicating the number of lists.
For each following K lines, output each list in lexicographical order.
For each list, output its length followed by the indexes in it separated by a single space.

示例1

输入

复制

abab

输出

复制

2
2 0 2
2 1 3

示例2

输入

复制

deadbeef

输出

复制

8
1 0
1 1
1 2
1 3
1 4
1 5
1 6
1 7

题意：

这个题意挺不好理解，看来很久（大概是我太菜了）大概是说给你一个串，然后从第一个字母开始向末尾添加，形成一个新串，如果这个新串和以前的旧串有相同，则分为一组，编号就是这个串是第几次首字母向末尾添加，如原串就应当是0。

题解：

kMP找出循环节，当我们找到循环节时，就可以确定在第几次添加时，是一个出现过的串。

举个例子；串 abbabb

　　　　　0:　　abbabb

　　　　　1:　　bbabba

　　　　　2:　　babbab

　　　　　3:　　abbabb

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int MAXN=1e6+10;
 4 char str[MAXN];
 5 int NEXT[MAXN];
 6 int getNEXT(int l)
 7 {
 8     NEXT[0]=-1;
 9     int k=-1;
10     int j=0;
11     while (j<l)
12     {
13         if(k==-1||str[j]==str[k])
14         {
15             ++k;
16             ++j;
17             NEXT[j]=k;
18         } else{
19             k=NEXT[k];
20         }
21     }
22 }
23 int main()
24 {
25     scanf("%s",str);
26     int len=strlen(str);
27     getNEXT(len);
28     int l=len-NEXT[len];//循环节的长度;
29     if(len%l!=0)
30     {
31         printf("%d\n",len);
32         for (int i = 0; i <len ; ++i) {
33             printf("%d %d\n",1,i);
34         }
35 
36     } else
37     {
38         printf("%d\n",l);
39         for (int i = 0; i <l ; ++i) {
40             printf("%d",len/l);
41             for (int j = i; j <len ; j+=l) {
42                 printf(" %d",j);
43             }
44             printf("\n");
45         }
46     }
47 
48 
49 
50     return 0;
51 }

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