2752 Seek the Name, Seek the Fame KMP 找出字符串中所有前缀-后缀的字符串

最新推荐文章于 2020-11-11 11:25:33 发布

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本文介绍了一种用于寻找字符串中既是前缀也是后缀的子串算法，并提供了完整的C++实现代码。该算法适用于帮助解决特定场景下名字生成的问题，如为新生儿创造既包含父名又包含母名的新名字。

Seek the Name, Seek the Fame

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5391		Accepted: 2381

Description

The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names to their newly-born babies. They seek the name, and at the same time seek the fame. In order to escape from such boring job, the innovative little cat works out an easy but fantastic algorithm:

Step1. Connect the father's name and the mother's name, to a new string S.
Step2. Find a proper prefix-suffix string of S (which is not only the prefix, but also the suffix of S).

Example: Father='ala', Mother='la', we have S = 'ala'+'la' = 'alala'. Potential prefix-suffix strings of S are {'a', 'ala', 'alala'}. Given the string S, could you help the little cat to write a program to calculate the length of possible prefix-suffix strings of S? (He might thank you by giving your baby a name:)

Input

The input contains a number of test cases. Each test case occupies a single line that contains the string S described above.

Restrictions: Only lowercase letters may appear in the input. 1 <= Length of S <= 400000.

Output

For each test case, output a single line with integer numbers in increasing order, denoting the possible length of the new baby's name.

Sample Input

ababcababababcabab
aaaaa

Sample Output

2 4 9 18
1 2 3 4 5

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
const int maxn=2000000;
int next[maxn];
char a[maxn],b[maxn];
//s数组从0开始，但是next数组计算时是从1开始(next[i]--s[i-1]),计算结果也从1开始
void get_next(char *s,int next[])
{
    int len=strlen(s);
    int i=0,j=-1;
    next[0]=-1;
    while(i<len)
    {
        if(j==-1||s[i]==s[j]) i++,j++,next[i]=j;
        else j=next[j];
    }
}
int res[maxn];
int main()
{
    while(gets(b))
    {
        int len=strlen(b);
        get_next(b,next);
        int cnt=0;res[0]=len;
        for(int i=len;next[i];i=next[i]) res[++cnt]=next[i];
        printf("%d",res[cnt]);
        for(int i=cnt-1;i>=0;i--) printf(" %d",res[i]);printf("/n");
    }
    return 0;
}