POJ 2752 Seek the Name, Seek the Fame——————KMP next 数组的运用

迈不过友情╰ 2022-05-16 09:27 152阅读 0赞

**Seek the Name, Seek the Fame**

Language:Default

Seek the Name, Seek the Fame

<table> 
 <tbody> 
  <tr> 
   <td><b>Time Limit:</b> 2000MS</td> 
   <td></td> 
   <td><b>Memory Limit:</b> 65536K</td> 
  </tr> 
  <tr> 
   <td><b>Total Submissions:</b> 24077</td> 
   <td></td> 
   <td><b>Accepted:</b> 12558</td> 
  </tr> 
 </tbody> 
</table>

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

Source

POJ Monthly–2006.01.22,Zeyuan Zhu

--------------------

找公共前后缀的最大长度  
比如第一个样列，  
*a**b**a**b**c**a**b**a**b**a**b**a**b**c**a**b**a**b*  a b a b c a b a b a b a b c a b a b   
公共前后缀有：

*  *a**b**a**b**c**a**b**a**b**a**b**a**b**c**a**b**a**b*  a b a b c a b a b a b a b c a b a b 
 *  *a**b**a**b**c**a**b**a**b*  a b a b c a b a b 
 *  *a**b**a**b*  a b a b 
 *  *a**b*  a b

长度分别是 18 9 4 2  
next数组记录的其实就是**最长公共前后缀的长度**

next\[i\]：字符串 0 到 i 的**最长公共前后缀的长度**

比如这个样列：  
*a**b**a**b**c**a**b**a**b**a**b**a**b**c**a**b**a**b*  a b a b c a b a b a b a b c a b a b

得到 next 数组之后，我们从 i=18 开始，  
*i*=18记录18*n**e**x**t*\[18\]=9  i = 18 记 录 18 n e x t \[ 18 \] = 9   
*i*=9记录9*n**e**x**t*\[9\]=4  i = 9 记 录 9 n e x t \[ 9 \] = 4   
*i*=4记录4*n**e**x**t*\[4\]=2  i = 4 记 录 4 n e x t \[ 4 \] = 2   
*i*=2记录2*n**e**x**t*\[2\]=0  i = 2 记 录 2 n e x t \[ 2 \] = 0   
*i*=0结束  i = 0 结 束

18,9,4,2  18 , 9 , 4 , 2

答案就是2 4 9 18

--------------------

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<stack>
    #include<iostream>
    using namespace std;
    const int MAXN=400000+7;
    char s[MAXN];
    int slen,nex[MAXN];
    
    void GetNext()
    {
        int i,j;
        i=0;
        j=nex[0]=-1;
        while(i<slen)
        {
            while(-1!=j && s[i] != s[j])    j=nex[j];
            nex[++i]=++j;
        }
    }
    int main()
    {
        while(~scanf("%s", s))
        {
            slen=strlen(s);
            GetNext();
            stack<int> sta;
    //        for(int i=0;i<=slen;i++)
    //            printf("next[%2d]:%2d\n",i,nex[i]);
            for(int i=slen;~i;i=nex[i])
                if(i>0)
                {
    //                printf("next[%2d]:%2d\n",i,nex[i]);
                    sta.push(i);
                }
            while(sta.size())
            {
                printf("%d ",sta.top());
                sta.pop();
            }
            printf("\n");
        }
        return 0;
    }