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

#include <bits/stdc++.h>
using namespace std;
const int MAXN=1e6+10;
char str[MAXN];
int NEXT[MAXN];
int getNEXT(int l)
{
    NEXT[0]=-1;
    int k=-1;
    int j=0;
    while (j<l)
    {
        if(k==-1||str[j]==str[k])
        {
            ++k;
            ++j;
            NEXT[j]=k;
        } else{
            k=NEXT[k];
        }
    }
}
int main()
{
    scanf("%s",str);
    int len=strlen(str);
    getNEXT(len);
    int l=len-NEXT[len];//循环节的长度;
    if(len%l!=0)
    {
        printf("%d\n",len);
        for (int i = 0; i <len ; ++i) {
            printf("%d %d\n",1,i);
        }

    } else
    {
        printf("%d\n",l);
        for (int i = 0; i <l ; ++i) {
            printf("%d",len/l);
            for (int j = i; j <len ; j+=l) {
                printf(" %d",j);
            }
            printf("\n");
        }
    }



    return 0;
}