Codeforces 432D Prefixes and Suffixes (next数组的应用)

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

题目链接 :http://codeforces.com/problemset/problem/432/D

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，子串长度要求增序输出

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[i]] += re[i];

#include <cstdio>
#include <cstring>
int const MAX = 1e5 + 5;
char s[MAX];
int next[MAX];
int len, pos, num;
int l[MAX], cnt[MAX], re[MAX];

void get_next()
{
int i = 0, j = -1;
next[0] = -1;
while(s[i] != '\0')
{
if(j == -1 || s[i] == s[j])
{
i ++;
j ++;
next[i] = j;
}
else
j = next[j];
}
}

void solve()
{
for(int i = 0; i <= len; i++)
re[i] = 1;
for(int i = len; i >= 1; i --)
if(next[i] != -1)
re[next[i]] += re[i];
int pos = len;
while(pos)
{
cnt[num] = re[pos];
l[num ++] = pos;
pos = next[pos];
}
}

int main()
{
scanf("%s", s);
len = strlen(s);
get_next();
solve();
printf("%d\n", num);
for(int i = num - 1; i >= 0; i--)
printf("%d %d\n", l[i], cnt[i]);
}