POJ 2752 Seek the Name, Seek the Fame

Seek the Name, Seek the Fame

Time Limit: 2000MS
Memory Limit: 65536KTotal Submissions: 11912
Accepted: 5844

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,


题目链接  :http://poj.org/problem?id=2752

题目大意  :给一个字符串，求其相同的前缀和后缀子串的串长

题目分析  :用到KMP算法中的next数组，本题时next数组的一种新用法，要深刻理解next数组的含义，由KMP算法的原理可知，通过遍历next[len], next[next[len]].....可以得到所有同时满足是原串的前缀和后缀的子串长度.这里举样例1具体说明

下标  :  0   1   2   3   4   5   6   7   8   9   10   11   12   13   14   15   16   17   18  
原串  :  a   b   a   b   c   a   b   a   b   a    b     a     b     c     a     b     a     b
next  : -1   0   0   1   2   0   1   2   3   4    3     4     3     4     5     6     7     8     9

串长18肯定是一个解，到18失配，next[18] = 9第二个解为9, 到9失配，next[9] = 4第三个解为4，到4失配，next[4] = 2第四个解为2
到2失配next[2] = 0结束，所以解集是{18,9,4,2}然后再逆序输出，这样做的原因是到18失配了，next值为9，所以原串的前9个字符和后9个字符必然是相同的(ababcabab)，那么我们只需要再看前9个字符组成的子串，next[9]＝4，说明前9个字符的前4个和后4个是相同的(abab)又因为后9个字符和前9个字符是相同的，所有后9个字符的前4个和后4个是相同的(abab)，因此我们可以推出前九个字符的前4个字符和后9个字符的后4个字符是相同的，即原串的前4个字符和后4个字符是相同的，2的情况一样，讲到这里相信那些对kmp很怵的童鞋也都能看懂了吧。

#include <cstdio>#include <cstring>int const MAX = 400000 + 5;char str[MAX];int next[MAX];int ans[MAX];void cal(){    next[0] = -1;    int k = -1, j = 0;    while(str[j] != '\0')    {        if(k == -1 || str[j] == str[k])        {            k++;            j++;            next[j] = k;        }        else            k = next[k];    }}int main(){    int count;    while(scanf("%s", str) != EOF)    {        count = 0;        cal();        int temp = strlen(str);        while(temp != 0)        {            ans[count++] = temp;            temp = next[temp];        }        for(int i = count - 1; i >= 0; i--)  //逆序输出            printf("%d ",ans[i]);        printf("\n");    }}