POJ 2752 Seek the Name,Seek the Fame

 

Seek the Name, Seek the Fame

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14048		Accepted: 6990

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

题目传送门：POJ 2752 Seek the Name,Seek the Fame

题意：输入一个字符串，找出该字符串的所有前缀的长度，使得该长度的前缀==后缀。说起来挺麻烦的- -其实看一下样例输入输出就能看出来了~

这是一道关于KMP next数组理解的题。设输入的字符串s的长度为len，s本身必满足条件。其他满足条件的子串有这么一个特征：子串的最后一个字符肯定==s串的最后一个字符。（和循环节的意思差不多），那么此时我们就可以运用next数组的性质。如下图：

下标：      0  1  2  3  4  5  6  7  8  9  10 11 12 13 14 15 16 17

串：          a  b  a  b  c  a  b  a   b  a   b   a   b    c   a   b    a   b

next值：  -1  0  0  1  2  0  1  2  3  4   3   4   3    4   5   6    7   8   9

我们设m=len-next[len]，此时m=9。那么我们再看next[m]，也就是next[9]，等于4。是不是也是输出的结果呢？继续看next[m]，也就是next[4]，等于2.

所以我们就找出了规律。首先m=len-next[len]。接下来我们的答案就是m组成的数组，m=next[m]，直到m到0或者-1为止。

至于为什么会这样呢？就是对于next数组的理解问题。next数组是避免不必要的逐个查找比对，也就代表匹配不成功后向后滑动多少个单位。它也代表了循环节里某个字符在上一个循环节里面出现在母串的位置，我们正是利用了next数组的这个性质。

其实有时候KMP的题搞不懂了，或者自己对next数组理解的还不够透彻，没法灵活运用。我们不妨按照样例先把next数组求出来，就像上表这样对照着看，然后依据样例答案找找规律，一般就能解出来了~ 下面附AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define TEST cout<<"wwwwwwwwwww"<<endl
#define N 1000005
char a[N];
int nextt[N],ans[N];
int len;
void getNext()
{
    int i,j;
    i=0;
    j=-1;
    nextt[i]=j;
    while(i<len)
    {
        if(j==-1 || a[i]==a[j])
        {
            i++;
            j++;
            nextt[i]=j;
        }
        else
            j=nextt[j];
    }
    return ;
}

int main()
{
    int k,m;
    while(scanf("%s",a)!=EOF)
    {
        k=0;
        len=strlen(a);
        memset(ans,0,sizeof(ans));
        memset(nextt,0,sizeof(nextt));
        getNext();
        m=nextt[len];
        while(m!=-1 && m!=0)
        {
            ans[k++]=m;
            m=nextt[m];
        }
        for(int i=k-1;i>=0;i--)
            printf("%d ",ans[i]);
        printf("%d\n",len);
    }
    return 0;
}