牛客网牛客网暑期ACM多校训练营（第三场）E KMP

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.

$\texttt{"a-z"}$

).

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.

输入例子:

abab

输出例子:

-->

示例1

输入

abab

输出

示例2

输入

deadbeef

输出

题意给定一个字符串T 设长度为n 然由这个字符串产生出n个字符串s0...sn-1 si=Ti-Tn-1+T0-Ti-1 将si相同的分到一组 i 按照从小到大排序输出分组后的 i 序列

解析我们可以推导一下若T是不循环的那么没有相同的若是循环的最小循环节为len 那么共有len 组 i 就属于 i%len这一组

　　　KMP判断是否循环且找出最小循环节长度随便保存一下也可以哈希。。。

 #include <bits/stdc++.h>

 #define pb push_back

 #define mp make_pair

 #define fi first

 #define se second

 #define all(a) (a).begin(), (a).end()

 #define fillchar(a, x) memset(a, x, sizeof(a))

 #define huan printf("\n");

 #define debug(a,b) cout<<a<<" "<<b<<" ";

 using namespace std;

 typedef long long ll;

 const int maxn=1e6+,inf=0x3f3f3f3f;

 const ll mod=1e9+;

 char t[maxn];

 int _next[maxn];

 int tlen,slen;

 void getnext()

 {

     int j,k;

     j=,k=-,_next[]=-;

     while(j<=tlen)

         if(k==-||t[j]==t[k])

             _next[++j]=++k;

         else

             k=_next[k];

 }

 int main()

 {

     while(scanf("%s",t)!=EOF)

     {

         tlen=strlen(t);

         set<int> s[tlen+];

         getnext();

         if(tlen%(tlen-_next[tlen])==)

         {

             int len=tlen-_next[tlen];

             for(int i=;i<tlen;i++)

                 s[i%len].insert(i);

             printf("%d\n",len);

             for(int i=;i<len;i++)

             {

                 printf("%d",s[i].size());

                 for(auto it=s[i].begin();it!=s[i].end();it++)

                 {

                     printf(" %d",*it);

                 }

                 huan;

              }

         }

         else

         {

             printf("%d\n",tlen);

             for(int i=;i<tlen;i++)

                 printf("%d %d\n",,i);

         }

     }

 }