POJ - 2752 Seek the Name, Seek the Fame (KMP)

阳光穿透心脏的1/2处 2022-05-19 08:57 92阅读 0赞

#                            Seek the Name, Seek the Fame #

### 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

### 题意描述： ###

给定一个串，求它的子串的长度，子串满足既是主串的前缀，又是主串的后缀，

将所有满足要求的子串长度按照升序输出。

### 解题思路： ###

在网上看了题解才明白了题意，但不是很理解为什么这样算，知道next\[\]数组是存储子串的前缀最大匹配，它在每次j=next\[j-1\]时找到了与它相同的字符再继直到j等于0.

别人的详细题解：[http://972169909-qq-com.iteye.com/blog/1071548][http_972169909-qq-com.iteye.com_blog_1071548]

#include<stdio.h>
    #include<string.h>
    int next[1000010];
    int s[1000010]; 
    void get_next(char s[])
    {
    	int i,j,len;
    	len=strlen(s);
    	j=0;
    	i=1;
    	next[0]=0;
    	while(i<len)
    	{
    		if(j==0&&s[i]!=s[j])
    		{
    			next[i]=0;
    			i++;
    		}
    		else if(j>0&&s[i]!=s[j])
    			j=next[j-1];
    		else
    		{
    			next[i]=j+1;
    			i++;
    			j++;
    		}
    	}
    }
    int main()
    {
    	int n,t,i,len,j;
    	char a[1000010];
    	
    	t=0;
    	
    	while(gets(a)!=NULL)
    	{
    		get_next(a);
    		len=strlen(a);
    		i=1;
    		s[0]=len;
    		j=next[len-1];
    		while(j>0)
    		{
    			s[i++]=j;
    			j=next[j-1];
    		}
    		for(j=i-1;j>0;j--)
    			printf("%d ",s[j]);
    		printf("%d\n",s[0]);
    	}
    	return 0;
    }

[http_972169909-qq-com.iteye.com_blog_1071548]: http://972169909-qq-com.iteye.com/blog/1071548