POJ 2752 Seek the Name, Seek the Fame kmp失配函數next應用

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Seek the Name, Seek the Fame Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 12791 Accepted: 6304

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
給你一個字符串，求所有前綴等於後綴的情況。 next數組的應用，KMP求出next數組，每次去掉next[i]到i的一段字符，剩余字串仍滿足條件，直到找到頭為止。 next[i]！=0的，都是模式串的前綴和後綴相同的字符數。

//3892K	547MS
#include
#include
#define M 10007
char pattern[400007];
int next[400007],m,ans[400007];
void pre(int len)
{
    int i = 0, j = -1;
    next[0] = -1;
    while(i != len)
    {
        if(j == -1 || pattern[i] == pattern[j])
            next[++i] = ++j;
        else
            j = next[j];
    }
}
int main()
{
    while(scanf("%s",pattern)!=EOF)
    {
        memset(ans,0,sizeof(ans));
        memset(next,0,sizeof(next));
        m=strlen(pattern);
        pre(m);
        int k=0;
        for(int i=m; i!=0; )
        {
            ans[k++]=next[i];
            i=next[i];
        }
        for(int i=k-2; i>=0; i--)
            printf("%d ",ans[i]);
        printf("%d\n",m);
    }
    return 0;
}