POJ：2753-Seek the Name, Seek the Fame

Seek the Name, Seek the Fame

Time Limit: 2000MS Memory Limit: 65536K

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

---

• 题意就是问字符串s中有多少个长度不同的相同的前缀和后缀子串。

• 就是考察的一个KMP中的next数组，考察相同的前缀和后缀就是next数组，然后next数组跳转就可以了。

  next[i]的意义就是：前面长度为i的字串的【前缀和后缀的最大匹配长度】

---

#include<stdio.h>
#include<cstring>
#include<vector>
using namespace std;
const int maxn = 4e5+100;
char s[maxn];
int next[maxn];

void cal_next()
{
    int k = -1;
    next[0] = -1;
    int n = strlen(s);
    for(int i=1;i<n;i++)
    {
        while(k>-1 && s[i] != s[k+1])
            k = next[k];
        if(s[i] == s[k+1])
            k++;
        next[i] = k;
    }
}

int main()
{
    while(scanf("%s",s) != EOF)
    {
        cal_next();

        int len = strlen(s);
        vector<int>ve;
        int k = len-1;
        ve.push_back(len);
        while(next[k] != -1)
        {
            ve.push_back(next[k]+1);
            k = next[k];
        }
        int n = ve.size();
        for(int i=n-1;i>=0;i--)
        {
            printf("%d",ve[i]);
            if(i != 0)
                printf(" ");
        }
        printf("\n");
    }
    return 0;
}