POJ 2752 Seek the Name, Seek the Fame [kmp]

Seek the Name, Seek the Fame

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17898		Accepted: 9197

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

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题意：给定字符串S，问所有满足既是S的前缀，又是S的后缀的子串的长度

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若将i的父结点设为f[i]，那么会形成一棵树。
对于i的祖先j，一定满足S[1,j]=S[i-j+1,i]。并且满足S[1,j]=S[i-j+1,i]的j，一定是i的祖先。
本题求的就是S的所有祖先的长度。

也就是说，它的失配函数的相同前后缀一定也是它的相同前后缀(相同前后缀的相同前后缀)

注意|S|一定成立

又犯了数组大小的沙茶错误

//
//  main.cpp
//  poj3461
//
//  Created by Candy on 10/19/16.
//  Copyright ? 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=4e5+5;
int f[N],n;
char s[N];
void getFail(){
    f[1]=0;
    for(int i=2;i<=n;i++){
        int j=f[i-1];
        while(j&&s[i]!=s[j+1]) j=f[j];
        f[i]=s[i]==s[j+1]?j+1:0;
    }
}
int ans[N],m=0;
void sol(){
    m=0;
    getFail();
    int j=f[n];
    while(j){ans[++m]=j;j=f[j];}
    for(int i=m;i>=1;i--) printf("%d ",ans[i]);
    printf("%d
",n);
}
int main(){
    //freopen("in.txt","r",stdin);
    while(scanf("%s",s+1)!=EOF){
        n=strlen(s+1);
        sol();
    }
    return 0;
}