POJ-2752 再谈Seek the name,Seek the same(KMP)

Seek the Name, Seek the Fame

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17648		Accepted: 9054

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

 

 

嗯这是我第二次写这题，对KMP也有了进一步的认识

 

由于数组是从0开始存的，这有一个好处，next[ls]表示的就是后缀和前缀相同的长度，同样由于next[ls]表示的是一个长度，那么以next[ls]为下标，正好是前面的绿线的后一位，也就是说next[next[ls]]表示的是前面的绿线前缀后缀(图中用zt表示)相同的最大长度，根据next的性质前面绿线的前缀zt和后面绿线的后缀zt是一个东西，所以可以通过next[next[......next[ls]]]这样的迭代找出题目要求的所有值

 

#include "bits/stdc++.h"
using namespace std;
typedef long long LL;
const int MAX=1000005;
int ls;
int next[MAX],ans[MAX];
char s[MAX];
void get_next(){
    int i,j;
    next[0]=-1;
    i=0,j=-1;
    while (i<ls){
        if (j==-1 || s[i]==s[j]){
            i++,j++;
            next[i]=j;
        }
        else j=next[j];
    }
}
int main(){
    freopen ("seek.in","r",stdin);
    freopen ("seek.out","w",stdout);
    int i,j;
    while (~scanf("%s\n",s)){
        ls=strlen(s);
        get_next();
        ans[0]=0;
        ans[++ans[0]]=ls;
        i=next[ls];
        while (i!=-1){
            ans[++ans[0]]=i;
            i=next[i];
        }
        sort(ans+1,ans+ans[0]+1);
        for (i=2;i<=ans[0];i++)
         printf("%d ",ans[i]);
        printf("\n");
    }
    return 0;
}