Codeforces 432D Prefixes and Suffixes kmp

手动转田神的大作：http://blog.csdn.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];

 

 

#include <cstdio>
#include <cstring>
int const MAX = 1e5 + 2;
char str[MAX];
int next[MAX], re[MAX], le[MAX], sub[MAX];;
int len, count;

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

void kmp()
{
    get_next();
    len = strlen(str);
    memset(re,0,sizeof(re));
    for(int i = 1; i <= len; i++)
        re[i] = 1;
    int temp = len;
    for(; temp > 0; temp--)
        if(next[temp])
            re[ next[temp] ] += re[temp];
    count = 0;
    le[count] = len;
    sub[count++] = 1;
    len = next[len];
    while(len)
    {
        le[count] = len;
        sub[count++] = re[len];
        len = next[len];
    }
}

int main()
{
    while(scanf("%s", str) != EOF)
    {
        kmp();
        printf("%d\n",count);
        for(int i = count - 1; i >= 0; i--)
            printf("%d %d\n",le[i], sub[i]);
    }
}