poj 2752 Seek the Name, Seek the Fame(KMP需转换下思想)

Seek the Name, Seek the Fame

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10204		Accepted: 4921

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

题意：

给你一个字符串。要你找出既是前缀又是后缀的子串长度的可能值。按字典序输出。

思路：

转化下思路其实很简单。既然既是前缀又是后缀。直接用文本串构造一个失配数组。然后直接匹配文本串的文本末。即txt[n].n=strlen(n)。而长度就为f[j]。想一想就明白了。感觉思想还是蛮好的。

详细见代码：

#include <iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
char txt[400100];
int f[400100],ans[400100],cnt;
void getf(char *p)//得到失配数组
{
    int i,j,m=strlen(p);
    f[0]=f[1]=0;
    for(i=1;i<m;i++)
    {
        j=f[i];
        while(j&&p[j]!=p[i])
            j=f[j];
        f[i+1]=p[j]==p[i]?j+1:0;
    }
}
int main()
{
    int len,i,j;

    while(~scanf("%s",txt))
    {
        getf(txt);
        len=strlen(txt);
        cnt=0;
        j=len;
        while(f[j])
        {
            ans[cnt++]=f[j];
            j=f[j];
        }
        for(i=cnt-1;i>=0;i--)
        printf("%d ",ans[i]);
        printf("%d\n",len);//注意自己是自己的前后缀
    }
    return 0;
}