poj 2752 Seek the Name, Seek the Fame（KMP）

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Seek the Name, Seek the Fame

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11198		Accepted: 5457

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

題意：給你一個字符串，找出所有的前綴和後綴相等的子串,按小到大輸出這些子串的長度，kmp變形

深刻理解next函數的意義，這題就變得非常簡單。

代碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;

char s[400005];
int e[400005];

int next[400005];
void getnext(char *s,int *next,int len)
{
    int i,j;
    next[0]=-1;
    i=0;
    j=-1;
    while(i<len)
 {
    if(j==-1||s[j]==s[i])
    {
        i++;
        j++;
        next[i]=j;
    }
    else
      j=next[j];
 }
}
int main()
{
    int t,i,j,k,n;
   while(~scanf("%s",s))
  {
      t=strlen(s);
      memset(next,0,sizeof(next));
      getnext(s,next,t);
     e[0]=t;
     i=t;n=1;

     while(next[i]>0)
     {
        e[n++]=next[i];
        i=next[i];
     }
     for(i=n-1;i>=0;i--)
     {
         printf("%d",e[i]);
         if(i!=0)
             printf(" ");
     }
      printf("\n");

  }
   return  0;
}