POJ 2752 Seek the Name, Seek the Fame KMP Next[]

给出一个字符串，找出所有既是它前缀又是它后缀的字符串，输出它们的长度。

Seek the Name, Seek the Fame

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 15400 Accepted: 7796

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

ababcababababcababaaaaa

Sample Output

2 4 9 181 2 3 4 5

Source

POJ Monthly--2006.01.22,Zeyuan Zhu

解法：利用next[]+递归/循环 即可

/**========================================== *   This is a solution for ACM/ICPC problem * *   @source：poj 2752 Seek the Name, Seek the Fame *   @type:  dp *   @author: wust_ysk *   @blog:  http://blog.csdn.net/yskyskyer123 *   @email: [e-mail] *===========================================*/#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>using namespace std;typedef long long ll;const int INF =0x3f3f3f3f;const int maxm=400000    ;//const int maxV=12    ;char M[maxm+10];int nex[maxm+10];void getnex(){    int len=strlen(M);    nex[0]=nex[1]=0;    for(int i=1;i<len;i++)    {        int p=nex[i];        while(p&&M[p]!=M[i]) p=nex[p];        nex[i+1]=M[p]==M[i]?p+1:0;    }    vector<int >ve;    ve.push_back(len);    int x=nex[len];    while(x)    {        ve.push_back(x);        x=nex[x];    }    for(int i=ve.size()-1;i>=0;i--)    {        if(i!=ve.size()-1)  putchar(' ');        printf("%d",ve[i]);    }    putchar('\n');}int main(){    while(~scanf("%s",M))    {        getnex();    }   return 0;}