POJ-2752 Seek the Name, Seek the Fame(kmp中next数组的应用)

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H - Seek the Name, Seek the Fame

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status Practice POJ 2752

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

next数组的应用，在做这题之前本以为自己掌握了kmp算法的精髓，但是在拿到这道题的时候竟然还是被卡了几个小时，真是伤心欲绝，

于是再次重温了next数组的本质和性质。这里引用别人总结的知识点来讲解一下吧

毕竟自己还是讲不清楚。

给定一个字符串s，从小到大输出s中既是前缀又是后缀的子串的长度。

此题非常简单，借用KMP算法的next数组，设s的长度为n，则s串本身必定满足条件。其他满足条件的子串都有个特征，就是该子串的最后一个字符肯定与s的最后一个字符相同。这正是next数组发挥作用的时候。从n - 1位既最后一位开始回滚，若s[next[n-1]] == s[n-1]，则子串s[0,1,2,...,next[n-1]]是满足条件的子串。然后判断s[next[next[n-1]]] == s[n-1]是否成立，这样一直回滚，直到next[next[.....next[n-1]]] == -1为止。把答案从大到小存下来，再从小到大输出即可。

大致思路：
如左图，假设黑色线来代表字符串str，其长度是len，红色线的长度代表next[len]，根据next数组定义易得前缀的next[len]长度的子串和后缀next[len]长度的子串完全相同（也就是两条线所对应的位置）。我们再求出next[len]位置处的next值，也就是图中蓝线对应的长度。同样可以得到两个蓝线对应的子串肯定完全相同，又由于第二段蓝线属于左侧红线的后缀，所以又能得到它肯定也是整个字符串的后缀。

所以对于这道题，求出len处的next值，并递归的向下求出所有的next值，得到的就是答案。

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<limits.h>#include<math.h>#include<algorithm>using namespace std;char a[400005];int p[400005],n,m;int sum[400005];void print(){int i,j=0;p[1]=0;for(i=2;i<=n;i++){while(j>0 && a[j+1]!=a[i])j=p[j];if(a[j+1]==a[i])j++;p[i]=j;}int k=0;for(i=n;i!=0;){sum[++k]=p[i];i=p[i];}for(i=k-1;i>=0;i--)  if(sum[i]!=0)printf("%d ",sum[i]);printf("%d\n",n);}int  main(){int i;while(scanf("%s",a+1)!=EOF){n=strlen(a+1);print();}}

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