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  • 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 Si 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 Si. Si and Sj will be the same group if and only if Si=Sj.
    3. For each group, let Lj be the list of index i in non-decreasing order of Si in this group.
    4. Sort all the Lj 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
    ",len);
    32         for (int i = 0; i <len ; ++i) {
    33             printf("%d %d
    ",1,i);
    34         }
    35 
    36     } else
    37     {
    38         printf("%d
    ",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("
    ");
    45         }
    46     }
    47 
    48 
    49 
    50     return 0;
    51 }
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  • 原文地址:https://www.cnblogs.com/-xiangyang/p/9377577.html
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