Codeforces 432D Prefixes and Suffixes kmp

手动转田神的大作:http://blog.****.net/tc_to_top/article/details/38793973

D. Prefixes and Suffixes

time limit per test
1:second
memory limit per test:
256 megabytes
input:
standard input
output:
standard output

You have a string s = s1s2...s|s|, where |s| is the length of string s, and si its i-th character.

Let's introduce several definitions:

  • A substring s[i..j(1 ≤ i ≤ j ≤ |s|) of string s is string sisi + 1...sj.
  • The prefix of string s of length l (1 ≤ l ≤ |s|) is string s[1..l].
  • The suffix of string s of length l (1 ≤ l ≤ |s|) is string s[|s| - l + 1..|s|].

Your task is, for any prefix of string s which matches a suffix of string s, print the number of times it occurs in string s as a substring.

Input

The single line contains a sequence of characters s1s2...s|s(1 ≤ |s| ≤ 105) —  string s. The string only consists of uppercase English letters.

Output 

In the first line, print integer k (0 ≤ k ≤ |s|) — the number of prefixes that match a suffix of string s. Next print k lines, in each line print two integers lici. Numbers li ci mean that the prefix of the length li matches the suffix of length li and occurs in string s as a substring ci times. Print pairs li ci in the order of increasing li.

Sample test(s)
input
ABACABA
output
3
1 4
3 2
7 1
input
AAA
output
3
1 3
2 2
3 1
 
 
题目大意 :给一个字符串,求其前缀等于后缀的子串,输出子串的长度和其在原串中出现的次数,子串长度要求曾序输出
 
题目分析 :首先得到母串的next数组,next数组的含义是next[j]的值表示str[0...j-1](我的next[0]是-1)这个子串的前后缀匹配的最长长度,如样例1
index  0  1  2  3  4  5  6  7
str    A  B  A  C  A  B  A
next   -1 0  0  1  0  1  2  3
next[6] = 2即ABACAB这个子串的前后缀最长匹配是2(AB)
由此性质我们可以发现满足条件的子串即是next[next[len。。。]]不断往前递归直到为0,因为长的可能会包含短的,我们可以递归得到re数组(re记录的就是子串出现的次数)re数组的递归式为re[ next[temp] ] += re[temp];
 
 
 1 #include <cstdio>
 2 #include <cstring>
 3 int const MAX = 1e5 + 2;
 4 char str[MAX];
 5 int next[MAX], re[MAX], le[MAX], sub[MAX];;
 6 int len, count;
 7 
 8 void get_next()
 9 {
10     int i = 0, j = -1;
11     next[0] = -1;
12     while(str[i] != '