牛客多校 2018 第三场 E Sort String（KMP求循环节）

2018 年 08 月 05 日 • 阅读: 1438 • 字符串 • 阅读设置

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|≤ 106
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.

输入

abab

输出

2
2 0 2
2 1 3

输入

deadbeef

输出

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

链接

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

代码

E答案正确147140481043
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1e6 + 10;
int Next[maxn];
char P[maxn];

void getNext(int len) // 求解next数组
{
memset(Next, 0, sizeof(Next));
Next[0] = -1, Next[1] = 0;
for (int i = 1; i < len; ++i)
{
int j = Next[i];
while (j != -1 && P[i] != P[j])
j = Next[j];
Next[i+1] = j + 1; // 关键是得到next[len]
}
}

int main()
{
scanf("%s", P);
int len = strlen(P);
getNext(len);
int k = len - Next[len]; // 最小循环节
if (len % k == 0)
{
printf("%d\n", k);
for (int i = 0; i < k; ++i)
{
int lenk = len / k; // 总长度/周期=每组数量
printf("%d", lenk);
for (int j = i; j < len; j +=k)
printf(" %d", j);
printf("\n");
}
}
else
{
printf("%d\n", len);
for (int i = 0; i < len; ++i)
printf("1 %d\n", i);
}
return 0;
}

The end.
2018-08-05 星期日