poj 2752 Seek the Name, Seek the Fame

Seek the Name, Seek the Fame

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 12866		Accepted: 6346

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

题目大意：给定一个字符串，求其相同前缀后缀的长度，由小到大输出所有情况。

解题思路：用kmp算法的next数组。由分析可知，整串字符串的相同前缀后缀中，若前缀子串有相同的前缀后缀，则

前缀子串的后缀一定等于后缀子串的后缀，即整串字符串有更小的相同前缀后缀，以此类推，可由大到小得出所有结果。

代码如下：

#include <cstdio>
#include <cstring>
int const maxn=400010;
char s[maxn];
int next[maxn];
int a[maxn];  //int a[maxn]; 
int len;
void get_next()
{
    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 main()
{
    int cur;
    while(scanf("%s",s)!=EOF)
    {
        int i=0;
		memset(a,0,sizeof(a));
        len=strlen(s);
        cur=len;
        a[0]=len;   //初始前缀长度一定为字符串长度
        get_next();
        while(cur)
        {
           a[++i]=next[cur];  //  next[前缀长度]即为前缀子串的前缀长度，直到next[]值为0是结束循环
           cur=a[i];
		}
        for(int j=i-1;j>=0;j--)
            printf("%d ", a[j]);
		printf("\n");
    }
    return 0;
}